MCAT Physics Equations**
A 2 kg ball on a string is rotated about a circle of radius 10 m. The maximum tension allowed in the string is 50 N. What is the maximum speed of the ball?
Uniform circular motion - The tension = Fc Fc = m∨² ---- r Fc = centripetal force m = mass ∨ = speed r = radius of the circular path
While driving a vehicle at a constant velocity on a flat surface, the accelerator must be slightly depressed to overcome reistive forces. how does the amount of work done by the engine ( via the accelerator) compare to the amount of work done by resistance?
Work energy theorum Wnet = ∆K W = work K= kinetic energy If the velocity is constant then the ∆k must be constant. That makes ∆k 0, so then the net work must also be 0. Answer: The amount of positive work done by the engine is equal to the amount of negative work done by the resistance.
FInd the x- and y- components of the following vector V=10m/s θ= 30°
X = V cos θ Y = V sin θ
Find the resultant of V1 and V2. For addition For subtraction
- For addition- place the tail of B to A and then draw a vector line pointing tip to tail. - For subtraction-flip the vector being subtracted (B) pointing from tip to tail. -For each vector sperate it into its X and Y components -Add the X components of both vectors together to get Rx. -Add the y components of both vectors together to get Ry. -Find the magnitude of the resultant using R² = Rx² + Ry²
How to calculate instantaneous acceleration at t= 6s?
- Must be provided a graph of velocity vs time - Draw a tangent line on the graph where the line crosses t= 6s. - Pick any 2 points on the tangent line write out their x (t) and y (V) intercepts - Plug these intercepts into the acceleration equation. a = ∆ V ----- ∆ t V = velocity t = time - If slope is + then acceleration will be + and point in the same direction as velocity - If slope is - then acceleration will be - and point in the opposite direction of velocity. https://www.youtube.com/watch?time_continue=111&v=Ed58JZKiGEM
A block is fully submerged 3 inches below the surface of a fluid, but is not experiencing any acceleration. What can be said about the displaced volume of fluid and the buoyancy force?
Buoyancy The displaced volume = volume of the block Bouyancy force = weight of the block = weight of the displaced fluid The block and the fluid must have the same density
Calculate the dot product of vector a (10) and b (4) with a angle of 30°
C = IAI IBI cos θ (we multiple by cos to make sure they point in the same direction)
A 3µF capacitor is connected to a 4V battery. if a piece of ceramic having dielectric constant k = 2 is placed between the plates, find the new charge, capacitance, and voltage of the capacitor?
Dielectric in a circuit capacitor
The H2O molecule has a dipole moment of 6.18 x 10⁻³⁰ C m. Calcualte the electrical potential due to a water moelcule at a point 89nm away along the axis of the diplole. (note k = 8.99 x 10⁹)
Electrical potential of a dipole moment V = kp --- cosθ r² V = electrical potential p = dipole moment r² = distance along the dipole θ = degrees point is away from the axis of the dipole. Answer: 6 x 10⁻⁶ V
Find the gravitational force between an electron and a proton that are 10⁻¹¹ m apart. (note: mass of a proton = 1.67 x 10⁻²⁷ kg. mass of an electron = 9.11 x 10⁻³¹ kg)
Fg = G m₁ m₂ ------- r² G = 7 x 10⁻¹¹ (universal gravitational constant) r = distance between 2 objects
If the distance between 2 objects is halved, what would happen to the gravitational force they exibit on eachother?
Fg would be quadroupled. If distance decreases gravitational force increases Fg = G m₁ m₂ ------- r²
What is the resulting vector if we multiple the vector by 3 (a scalor) vector has a x = 2 and a y= 1
Magnitude -Multipy the 3 by each X and Y . Direction: -If scalor = + the resultant will point in the same direction as the origional vector. -If scalor = - the resultant will point in the opposite direction of the origional vector.
Suppose a proton is moving with a velocity of 15 m/s towards the top of the page through a uniform magnetic field of 3.0 T directed into the page, as shown on the PIC ON OTHER SIDE. What is the magnetiude and direction of the magnetic force on the proton? Describe the motion that will result from this setup. (note: the charge of a proton is 1.60 x 10 ⁻¹⁹ C, and its mass is 1.67 x 10⁻²⁷ kg.)
Magnetic force Magnitude of the magnetic force: FB = q V B sinθ FB = magnetic force q = point charge V = velocity of the charge B = magnetic field θ = the angle between the velocity and magnetic field vector Magnitude of the force: FB = 7.0 x 10⁻¹⁸ Fb direction: radially inward towards the center of the circle Motion that results: the proton will move in a circle with a radius of 50 x 10⁻⁹ m
What is the magnitude and direction of the resultant vectors from the cross product of C= A x B A: X = -3 N, Y = 0 B: X = 0, Y = +4 m
Magnitude C = IAI IBI sin θ Direction- use right hand rule A= thumb (1st vector) B= fingers (2nd vector) C= Palm Pt thumb to the left and fingers up= palm is into the page. Almost all physics probs use this.
If the magnetic field a distance r away from a current carrying wire is 10 T, what will be the net magnetic field at r if another wire is placed a distance 2r from the origional wire (with r in the middle) and has a current twice as strong flowing in the opposite direction?
Magnitude of magnetic field B = µ₀I --- 2r B = magnetic field µ₀ = permeability of free space (a constant) I = current r = how far the magnetic field is from the wire
A tired squirrel (mass of approximately 1 kg) does push-ups by applying a force to elevate its center-of-mass by 5 cm in order to do a mere 0.50 Joule of work. If the tired squirrel does all this work in 2 seconds, then determine its power.
Power P = W --- t P = Power W = Work (which is equal to ∆E, the change in energy) t= time 0.50 J/ 2s = 0.25 W Answer: 0.25 W
The window of a skyscraper measures 2.0m by 3.5 m. If a storm passes by and lowers the pressure outside the window to 0.997atm while the pressure inside the building remains at 1 atm, what is the net force pushing on the window.
Pressure There will be a net force pushing on the window in the direction of the lower pressure (outside the window) P = F / A Fnet = Pnet A Fnet = (Pinside - Poutside) A P = Pressure (in Pa) F = magnitude of the normal force vector A = area Must convert pressure into Pa ∆P = (1-0.997) = 0.003 atm 1 x 10⁵ Pa = 1atm so 1 x 10⁵ Pa/1atm x 0.003 atm = 3 x 10² Pa now multiply by area (3 x 10²) (2 x 3.5) Answer: 2100N
Where is the center of mass for a symmetric object?
The geometric center of the object.
What is the magnitude of the vector with the following components? x= 3 m/s y= 4 m/s
V² = X² + y²
An anchor made of iron weighs 833 N on the deck of a ship. If the anchor is now suspended in seawater by a massless chain, what is the tension in the chain? (Note the density of iron is 7800 and the density of seawater is 1025)
Boyancy force Answer: 724N
How to calculate the center of mass for an object that is asymetric? Like a person or TV
You must seperate them into x, y and Z coponents. The x's and y's are coordinates of the objects position on a graph so u have to be given a graph. x = m₁⋅×₁ + m₂⋅×₂ +....ect for each object ----------------- m₁ + m₂ + ....... y = m₁⋅y₁ + m₂⋅y₂ +.... ------------------ m₁ + m₂ + ....... https://www.youtube.com/watch?v=VrflZifKIuw&feature=youtu.be
Is μs or μk always larger than the other? What does this mean?
μs is always larger than μk. This means that it takes more force to get an object to start sliding then it does for a object to keep sliding.
A 15kg block, initally at rest slides down a frictionless incline and comes to the bottom with a speed of 7m/s, as shown below. What is the kinetic energy of the object at the top and bottom of the ramp?
Kinetic energy K = ½ m∨² K = kinetic energy m = mass ∨ = speed Top of ramp - when at the top of the ramp, the block is at rest and so has a speed of 0. - v = 0 - Answer: K= 0 J Bottom of ramp - Plug in the values - Answer: K = 375 J
A ball is thrown vertically up into the air from a window ledge 30 meters above the ground with an inital velocity of 10 m/s. Find the velocity and position of the ball after 2 seconds.
Linear motion For velocity V = V₀ + at V = Velocity final V₀ = Velocity inital a = acceleration (equal to -10 m/s²) t = time For Position (we use Y direction because it was thrown vertically and so X is not changing) y = V₀t + ½ at² y = position of the ball
A ball is thrown vertically upward with a velocity of 26.3m/s. What is the total time the ball was in flight?
Linear motion THE AMOUNT OF TIME IT TAKES AN OBJECT TO REACH MAX HEIGHT IS THE SAME AMOUNT OF TIME IT TAKES AN OBJECT TO FALL BACK DOWN TO STARTING HEIGHT. Time it takes to reach max height V = V₀ + at V = Final Velocity (0 at max height) V₀ = inital velocity a = acceleration t = time Total time in flight Take t from max height and multiple it by 2.
A ball is thrown vertically up into the air from a window ledge 30 meters above the ground with an inital velocity of 10 m/s. Find the distance and time at which the ball reaches its maximum height above the window ledge
Linear motion WHEN AN OBJECT IS AT ITS MAX HEIGHT IT V=0 Max height (We use Y only becuase it was thrown vertically) V² = V₀ + 2ay V = Final Velocity (which is 0 at max height) V₀ = Inital velocity a = acceleration y = max height Time V = V₀ + at V = 0 at max height
Suppose a wire of length 2.0 m is conducting a current of 5.0 A toward the top of the page and through a 30 gauss uniform magnetic field directed inot the page. what is the magnitude and direction of the magnetic force on the wire?
Magnetic Force on a current carrying wire FB = I L B sin θ FB = magnetic force I = current L = length of the wire B = magnetic field θ = angle between L and B Answer: FB = 3 x 10⁻² N FB points to the left
Consider a person on a sled sliding down a 100 m long hill on a 30° incline. The mass is 20 kg, and the person has a velocity of 2 m/s down the hill when they're at the top. How fast is the person traveling at the bottom of the hill?
Conservation of mechanical energy Because the only force acting on the person is force of gravity, which is a conservative force, we can use conservation of mechanical energy 0 = ∆E = ∆U + ∆K ∆E= total change in mechanical energy ∆U = total change in potential energy ∆K = total change in kinetic energy K = ½ m∨² K = kinetic energy m = mass ∨ = speed U = mgh U = potential energy m = mass g = acceleration due to gravity= 10 m/s² h = distance between the object and the ground level (usually the ground) This is the same thing as: ∆Ei = ∆Ef Ui + Ki = Uf + Kf hi = 100m ⋅ sin(30°) hf= 0 (becuase at the bottom) so u can eliminate Uf. m = 20kg vi = 2 m/s vf = ? New equation: Ui + Ki = Kf mgh + ½ m∨² = ½ m∨² (20)(10)(100m ⋅ sin(30°)) + ½ (20)(2)² = ½ (20)∨² 9840 J = ½ (20)∨² Answer: ∨ = 31.3 m/s
A 40kg block is resting at a height of 5m off the ground. If the block is released and falls to the ground, what is the total mechanical energy at a height of 2m, assuming negligable air resistance?
Conservation of mechanical energy Since air resistance is negligable, we can assume mechanical energy is constant as it falls. If we find what the mechanical energy is at the top, we can determine what it is at the bottom. 0 = ∆E = ∆U + ∆K ∆E= total change in mechanical energy ∆U = total change in potential energy ∆K = total change in kinetic energy Because at the starting height of 5m, there is no kinetic energy, we can say that E= ∆U U = mgh U = potential energy m = mass g = acceleration due to gravity= 10 m/s² h = distance between the object and the ground level (usually the ground) Answer: 2000 J
An electron is acclerated over a distance d by an electrical potential V. The electrical potential applied to this electron is then increased by a factor of 4 and the electron is accelerated over the same distance d. The speed of the electron at the end of the second trial will be larger than at the end of the first trial by a factor of?
Correct answer: increases by a factor of 2 V = W/ q V = electrical potential W = work q = point charge
A proton and an alpha particle (a helium nucleus) repel eachother with a force of F while they are 20nm apart. If each particle combines with 3 electrons, what is the magnitude of the new force between them?
Correct: F Fe = kq₁q₂ ------- r² (or kQq) Fe = electrostatic force k = Coulombs constant = 9 x 10⁹ q₁ and q₂ = magnitude of the 2 charges r = distance between 2 charges
A wooden block floats in the ocean with half its volume submerged. Find the density of the wood ρb. (note: the density of the sea water is 1025 kg/m³)
Bouyancy force Fbuoy = ρ(fluid) V(fluid displaced)g Fg = ρ(fluid) V(the object is submerged in)g Fbouy= boyance force ρ = density of the fluid V = Volume of fluid displaced g = gravity Fbouy = Fg because the object is floating half submerged. Vfluid displaced = Vb/2 Answer: 510 kg/m³
A block weighing 200 N is pushed along a surface. If it takes 80 N to get the block moving and 40 N to keep the block moving at a constant velocity, what are the coefficients of friction μs and μk?
ƒs = μs N (80 = μs 200) ƒs = static friction μs = coefficent of static friction N = Normal force ƒk = μk N (40 = μk 200) ƒk = kinetic friction μk = coefficent of kinetic friction N = Normal force
An office building with a bathroom 40m above the ground has its water supply enter the building at ground level through a pipe with an inner diameter of 4cm. If the linear speed at the ground floor is 2 m/s and at the bathroom is 8m/s, determine the cross-sectional area of the pipe in the bathroom. If the pressure in the bathroom is 3 x 10⁵ Pa, what is the required pressure at ground level?
Flow through a pipe. -Continuity Equation and Bernoullis equation For Cross sectional area: Continuity equation Q = v₁A₁ = v₂A₂ Q = flow rate v₁ and v₂ = linear speed of the fluids at points 1 and 2. A₁ and A₂ = cross sectional areas at points v₁ and v₂ Area of a circle: A = π r² For Pressure: Bernoullis equation -Since the height at the ground level is 0 u can cancel out of pgh₁ term P₁ + ρgh₁ + ½ρ∨₁² = P₂ + ρgh₂ + ½ρ∨₂² P = Absolute pressure at the location of the pipe. ρ = density of the fluid g = acceleration due to gravity (g) h= height of the fluid above some datum ∨ = linear speed Area Answer: 3.0 x 10⁻¹⁴ m² Pressure Answer: 7.3 x 10⁵ Pa
Objects A and B are submerged at a depth of 1m in a liquid with a specific gravity of 0.877. Given that the density of the object B is one-third that of object A and that thee gauge pressure of object A is 3atm, what is the gauge pressure of object B?
Gauge Pressure Answer: 3atm The absolute and gauge pressure depend on only the density of the fluid, not the object. Pgauge = ρgz ρ = density of the fluid g = acceleration due to gravity (10m/s²) z = depth of the object Pgauge = 1000 x 10 x 20 Since they are at the same depth, in the same fluid, then they must have the same gauge pressure.
A driver in the ocean is 20m below the surface. What is the gauge pressure at her depth? What is the absolute pressure she experiences? (the density of sea water is 1025 kg/m³)
Gauge Pressure and Atmospheric Pressure Gauge Pressure - Since the pressure at the surface( P₀) is the same as the pressure in the atmosphere (Patm) we can cancel out P₀ and Patm from the normal equation. We are left with: Pgauge = ρgz ρ = density of the fluid g = acceleration due to gravity (10m/s²) z = depth of the object Pgauge = 1000 x 10 x 20 Absolute Pressure P = Patm + Pgague P = absolute pressure Patm = atmospheric pressure P = 1 x 10⁵ Pa + 2 x 10⁵ Pa (use 1 x 10⁵ becuase thats the value for atmospheric pressure u memorize) Answer: Pgauge = 2 x 10⁵ Pa Answer: Patm = 3 x 10⁵ Pa
A large cylinder is filled with equal volumes of 2 immiscible fluids. A balloon is submerged in the first fluid: the guage pressure in the balloon at the deepest point in the first fluid is found to be 3atm. Next the balloon is lowered all the way to the bottom of the second fluid, where the hydrostatic pressure in teh balloon reads 8 atm. what is the ratio of the guage pressure accounted for by the first fluid to the gauge pressure accounted for by the second fluid.
Gauge pressure Answer: 3:5
A 80kg diver leaps from a 10m cliff into the sea. Find the divers potential energy at the top of the cliff and when he is 2 meters underwater, using sea level as the datum.
Gravitational potential energy U = mgh U = potential energy m = mass g = acceleration due to gravity= 10 m/s² h = distance between the object and the ground level (usually the ground) Top of the cliff - h = 10 m because sealevel is the datum - Plug in - Answer: U = 8000 J 2m under water - h = -2 m beucase sealevel is datum - Plug in - Answer: U = -1600 J (yes K can be negative)
A 5kg block slides down a frictionless incline at 30°. Find the normal force and acceleration of the block.
Inclined planes- Motion in 2 dimensions -Draw the incline with the box on it -Draw all the forces acting on the block (Fg and N) (N is always perpendicular to the plane) -Split gravity into parallel and perpendicular forces (essentially X and Y) FgII = mgsinθ Fg⊥ = mgcosθ FgII = force of gravity parallel to the plane. Fg⊥ = force of gravity perpendicular to the plane Normal Force Fg⊥ = N becuase N is perpendicular to the plane and there is no acceleration in the perpendicular direction mgcosθ = N Acceleration - Newtons second law: Fnet = ma - Since we are only accelerating in the parallel direction, we are only concerned with the acceleration in the parallel direction so we only care about FnetII. -Therefore: FnetII = FgII -Therefore: FgII = maII mgsinθ = maII - Cancel out masses - gsinθ = aII
Suppose a wire is formed into a loop that carries a current of 0.25 A in a cloclwise direction (LOOK AT PIC TO SEE). Determine the direction of the magnetic field produced by this loop within the loop and outside the loop. If the loop has a diameter of 1m, what is the magnitude of the magnetic field at the center of the loop? (note: µ₀ = 12 x 10⁻⁷ )
Magnetic fields through a circular wire Direction of magnetic field thumb = current magnetic field = fingers curl around the wire - within the loop- fingers point into the page = magnetic field inot the page - outside of the loop- fingers point out of the page= magnetic field outside of the page Magnitude of magnetic field B = µ₀I --- 2r B = magnetic field µ₀ = permeability of free space (a constant) I = current r = how far the magnetic field is from the wire B = (12 x 10⁻⁷) (0.25A) ---------------- 2 x 0.5 B = 3.1 x 10⁻⁷ T
A projectile is fired from ground level with an inital velocity of 50 m/s and an inital angle of elevation of 37°. Find the projectiles total time in flight Find the total horizontel distance traveled
Projectile motion Time in flight THE AMOUNT OF TIME IT TAKES AN OBJECT TO REACH MAX HEIGHT IS THE SAME AMOUNT OF TIME IT TAKES AN OBJECT TO FALL BACK DOWN TO STARTING HEIGHT. So we use the y component of velocity. - V₀y = V₀ sinθ - Plug V₀y into the following equaiton to solve for t y = V₀y t + ½ ay t² y= 0 at inital height. Horizontal distance We only use the X component since we are only concerned with horizontal distance traveled. THE TIME IS THE SAME IN BOTH THE X AND Y DIRECTION. THE HORIZONTAL VELOCITY IS CONSTANT BECUASE THERE ARE NO FORCES ACTING IN X DIRECTION (U CAN ALWAYS ASSUME THIS). THERE IS ONLY ACCELERATION IN THE VERTICAL DIRECITON SO ax = 0. - V₀x = V₀ cosθ - Plug V₀x into the followign equation to solve for x. x = V₀x t + ½ ax t² ax = 0 becuase there is only acceleration in the vertical direction.
A seasaw with a mass of 5kg has one block of mass 10kg two meters to the left of the fulcrum and another block 0.5 m to the right of the fulcrum. If the seasaw is in equilibrium, find the mass of block 2 and the force exerted by the fulcrum.
Rotational equillibrium (because has a fulcrum, but is not rotating since its in equillibrium) Mass of block 2 - When in equillibrium, the torque on block 1 must be equal and opposite to the torque on block 2. τ₁ = τ₂ τ₁ = r₁g₁ sinθ₁ τ₂ = r₂g₂ sinθ₂ - Becuase the fulcrum is centered under the seesaw, both the normal force and weight of the seasaw will be eliminated from the eq becuase thier lever arms are 0. - solve by setting the torques equal to eachother. Normal force exerted by the fulcrum -Free body diagram shows N = Fg + seasaw + blocks -N = (mseesaw + m₁ + m₂) g
A circuit is wired with one cell supplying 5V in series with 3 resisitors of R1= 3Ω, R2= 5Ω, R3= 7Ω, wired in series. What is the resulting voltage across and current through each resistor of this circuit as well as the entire circuit?
Series circuits - For resistance Rs = R₁ + R₂ + R₃ - For current, the current is the same throughout the entire circuit. V = IR V = Voltage I = current R = resistance - To find Voltage drop across each resistor V = IR Answer: The current across each resistor is 0.33 A Answer: V1 = 1V V2 = 1.67V V3 = 2.3 V
Find the specific gravity of benzene, given that its density is 877 kg/ m³
Specific gravity - it is simply the density of the fluid over the density of water. - density of water is 1g/cm³ or 1000 kg/m³ SG = ρ / density of water SG= specific gravity ρ = density of a fluid Answer: 0.877
In the diagram below (SEE PIC ON OTHER SIDE), an electron goes from point a to point b in the vicinity of a very large positive charge. The electron could be made to follow any of the paths shown. Which path requires the least work to get the electron charge from a to b?
They all require the same amount of work becuase work is independent of the path taken to get there since they are on equipotential lines.
2 blocks are in equilibrium on a pully system. One box is on a table and the other is hanging off the table. If block A has a mass of 15 kg and the coefficent of static friction between block A and the surface is 0.2, what is the max mass of block B?
Translational equillibrium Make a free body diagram of each block with all the forces acting on it. Block A - T points to the right - Fs points to the left Block B - T points up From the free body diagram we can derive what the forces are equal too, which can tell us the mass. - ƒs = T - T = FgB Sub these 2 into equations eachother to get ƒs = FgB Since its asking for max mass of block B, we know we need to have ƒs mazimized, which is equal to: ƒs = μs N Plug ƒs equation into ƒs = FgB to solve.
Two parallel conducting plates are seperated by a distance d. One plate carries a charge + Q and the other carries a charge -Q. The voltage between the plates is 12V. IF a +2 uC charge is released from rest at the positive plate, hwo much kinetic energy does it have when it reaches the negative plate?
U = q ∆V U = electrical potential energy q = point charge V= electrical potential
A 2 kg ball on a string is rotated about a circle of radius 10 m. The maximum tension allowed in the string is 50 N. What is the velocity of the ball?
Uniform circular motion Velocity would be the force tangental to the object at any point in time (instaneous velocity) The velocity is 0 at the point in time because the speed in uniform circular motion is constant so there is no acceleration at any point in time.
The voltage across the terminals of an isolated 3µF capacitor is 4V. If a piece of ceramic having a dielectric constnat k = 2 is placed between the plates, find the new charge, capacitance and voltage of the capacitor
Dielectric in an isolated capacitor C = Q/ V C = Capacitance Q = Charge V = potential difference/ voltage
If the average speed of earth in one year is 30 km/s, what is its average velocity?
V = 0 km/s becuase Velocity = Δ x / Δ t. Δ x = 0 and so V = -0
Equations for Linear Motion
V = V₀ + at x = V₀t + ½ at² V² = V₀² + 2ax x = v̅t x = displacment V = Final Velocity V₀ = Inital velocity a = acceleration t = time v̅ = average velocity (just 1 squared in each equation)
A weight lifter lifts a 275 kg barbell from the ground to a height of 2.4 m. How much work has he done in lifting the barbell, and how much work is required to hold the weight at that height?
Work W = F ⋅ d ⋅ cosθ W = work F = applied force d= displacement through force is applied θ = angle between applied force and displacement vector lifting the barbell - since he lifts it parallel to the displacement, the angle is 0 cos (0) = 1 so W = F ⋅ d holding the weight - because there is no change in height d= 0 so no work is done Lifting the barbell: Answer: 7000 J holding the weight: Answer: 0 J
A system of 6 pullys has an efficiency of 80 percent. A person is lifting a mass of 200kg with the pully. Find: A. The distance through which the effort must move to raise the load a distance of 4m. B. The effort required to lift the load. C. The work done by the person lifting the load through a height of 4 m
Efficiency of a simple machine Efficiency = W out (load)(load distance) ----- = -------------------- W in (effort)(effort distance) - Wout = work that comes out of the system (displacement of mass to some height) - Win = work put into the system (exertion of force through a distance of rope) - load- weight of the object (mg) - load distance- height the object is lifted in the air - effort-force required to lift the object - effort distance- how much rope is pulled to get the object to where it needs to be. A. Find effort distance Since there are 6 pully and the object moves 4m, we know that each one gets moved 4m for a total effort distance of (4 x 6)= 24m. B. Find effort Efficiency = (load)(load distance) -------------------- (effort)(effort distance) efficiency - 80/100--> 0.8 load- mg- 200kg x 10---> 2000 load distance-4m effort distance-24m (calculated in A) effort- solve for C. Find Work in W in = (effort)(effort distance) effort- use from B effort distance- use from A A. Answer: effort distance= 24m B. Answer: Effort = 400N C. Answer: Work in = 9600 J
Two carts connected by a 0.50 m spring hit a wall, compressing the spring to 0.25m. The spring constant k is 200 N/m. What is the elastic potential energy stored from the spring's compression?
Elastic potential energy U = ½ kx² U = potential energy k = spring constant (a measure of the stiffness of the spring) x = how far it is stretched above or below its equilibrium length x= 0.25 -0.50 x = - 0.25m Now plug into equation Answer: U = 6 J (0.25^2 = 0.06. This isnt a MCAT prob, so if u didnt get that right dont worry about it) (prob dont need to be able to do that)
If the mass of one object is trippled, what would happen to the graviatation force it exhibits on another object?
Fg would be trippled mass is directly related to Fg. Fg = G m₁ m₂ ------- r²
An ideal gas in a sealed container is taken through the process shown in the PV diagram below. The initial volume of the gas is Vi = 0.25m³ and the final volume of the gas is Vf = 0.75m³. The inital pressure of the gass is Pi = 70,000 Pa and the final pressure of the gas is Pf = 160,000 Pa. What is the work done on the gas during the process shown? See pic on other side to solve the problem
Work done by a gas. -W done by the gas = area under the curve - Since the area under the curve is not a perfect rectangle, we must divide it into a rectangle and square. W = area₁ + area₂ Area of triangle = ½ ∆P ⋅ ∆V Area of a rectangle = P⋅∆V P = pressure= height (y) V = volume= base (x) To solve for the rectangle: Area₁ = Pi ⋅ ∆V (use Pi because thats the height) (70,000) ⋅ (0.75 - 0.25) Area₁ = 35,000 J To solve for the triangle: Area₂ = ½ ∆V ⋅ ∆P ½ (0.75- 0.25)⋅(16000 - 70,000) Area₂ = 22,500 J Total W: Area₁ + Area₂ Answer: 57,500 J
A lead ball of mass 0.125 kg is thrown stright up in the air with an inital velocity of 30 m/s. Assuming no air resistance, find the work done by the force of gravity by the time the ball is at its maximum height.
Work energy theorum Since all we know is the mass and velocity, its hard to use the regular work equation (W = Fd cosθ) Instead, since we know the net work is equal to the change in kinetic energy, we can just find K Wnet = ∆K W = work K= kinetic energy K = ½ m∨² K = kinetic energy m = mass ∨ = speed Since at max height the velocity is 0, Kf= 0 W = 0 - Ki 0 - ½ (0.125)(30²) Answer: W = -60 J
What is the displacement and distance of a man who walks 2km east, then 2km north, then 2km west and then 2km south?
distance= 8km (how much he walked, doesnt matter direction) displacement= 0km (the difference between his inital and final position) (He ends up the same place he started)
Two wooden balls of equal volume but different density are held beneath the surface of a container of water. Ball A has a density of 0.5 and ball B has a density of 0.7. When the balls are released they will accelerate upward to the surface. What is the releationship between the acceleration of ball A and that of ball B? A. Ball A has the greater acceleration B. Ball B has the greater acceleration C. Balls A and B have the same acceleration D. It cannot be determined from the information given.
Answer: A. Ball A has the greater acceleration Both balls expereince the same Fbouyancy becuase they are in the same liquid and displace the same volume. The ball with the smaller mass has the greater acceleration. Since ball A has a smaller density (m = pv), it will also have a smaller mass.
A block weighing 100 N is pushed up a frictionless incline over a distance of 20m to a height of 10m. Find: A. The minimum force required to push the block B. The work done by the force C. The force required and the work doen by the force if the block were simply lifted vertically 10m. Pic on other side. Need to look at it to solve the prob.
Mechanical advantage- simple machines- inclined planes A. minimum force - Anytime u see a inclined plane, split the force of gravity vector into parallel and perpendicular components. - Draw a free body diagram and include the Fn, Fapplied, Fg⊥, fg‖ - Fn= Fg⊥ so we can cross those out. - as a result Fapplied = Fg‖. Fg‖ = mg sin θ Fg‖ = parallel component of gravity mg = weight of the object (mass + gravity)= 100 N θ = angle between Fg and F which is the angle on the triangle. sin = ratio of opp/hyp--> 10/20 So, Fapplied=Fg‖= 100N (10/20) B. Work done by the force W = F ⋅ d ⋅ cosθ W = work F = applied force d= displacement through force is applied θ = angle between applied force and displacement vector - plug in the F from the last problem. - Since F and d are both parallel (because we are moving the block parallel) the angle between them is 0. cos(0) = 1 - so, W = 50N ⋅ 10m C. Work done by the force if lifted We know that mechanical advantage tools do not change the amount of work done, just the force needed. Beucase of this, we know the work done without it will be the same. W = F ⋅ d ⋅ cosθ W = 100N ⋅ 10m A. Answer: Fapplied= 50 N B. Answer: W = 1000 J C. Answer: W = 1000 J
A baseball of mass 0.25kg is thrown in the air with an inital speed of 30 m/s, but becuase of air resistance, the ball returns to the ground with a speed of 27 m/s. Find the work done by air resistance.
Noncconservative total mechanical energy (air resistance is a nonconservative force) W (nonconservative) = ∆E = ∆U + ∆K ∆E= total change in mechanical energy ∆U = total change in potential energy ∆K = total change in kinetic energy Since the ball returns to the ground (implying thats where it started) the potenital energy will be conseerved and the same so ∆U = 0 Since the speed changes, the kinetic energy is not conserved, so to find the total energy done by air resistance we just need to solve for ∆K W(nonconservative)= ∆E = ∆K ∆K = (½ m∨²) - (½ m∨²) K = kinetic energy m = mass ∨ = speed Answer: -20 J (its negative becuase energy is being lost from the system)
Consider 2 resistors wired in parallel with R1 = 5Ω and R2 = 10Ω. If the voltage across them is 10V what is the current through each of the 2 resistors?
Parallel circuits - For resistance 1 1 1 1 -- = --- + --- + -- Rp R₁ R₂ R₃ - For voltage, the voltage of the source is the same as the branch Vp = V₁ = V₂ = V₃ - To find current: use Ohms law V = IR V = voltage I = current R = resistance Answer: I₁ = 2A I₂ = 1 A
A hydraulic press has a piston of radius 5cm, which pushes down on an enclosed fluid. A 50 kg weight rests on this piston. Another piston in contact with this system has a radius of 20cm. Taking g =10m/s², what force is needed on the larger piston to keep the press in equilibrium?
Pascals Principle and hydraulic systems Pressure on one side is equal to the pressure on the other side. P₁ = P₂ F₁/A₁ = F₂/A₂ F = Applied force (mg) A = Area A= A = π r² A = area r = radius π = 3.0 Set up a simple ratio and solve F₁/A₁ = F₂/A₂ (500 x 1,200) / 75 Answer: 8000N
