Chapter 6 Practice

A spring has a spring stiffness constant k of 70.0 N/m How much must this spring be compressed to store 40.0 J of potential energy?

1/2*k*x²= energy stored in J AWS: x= 1.07 m

A 285-kgkg load is lifted 24.0 mm vertically with an acceleration a=0.110 gg by a single cable. Part A: Determine the tension in the cable. PART B: Determine the net work done on the load. PART C: Determine the work done by the cable on the load. PART D: Determine the work done by gravity on the load. PART E: Determine the final speed of the load assuming it started from rest

A) T-W= ma T=w+ma T=m(g+a) AWS: Ft= 3100 N B) W=Fd W=(T-W)d W=(ma)d AWS: 7370 J C) W= Td cos (theta) AWS: Wcable = 7.44×104 J D) the displacement is upwards and the weight points downwards, so the angle is 180º W = F. d W = F d cos 180 AWS: WG = −6.70×104 J E) W = ΔK as part of rest K₀ = 0 W = ½ m v_f² AWS: v= 7.19 m/s

A roller-coaster car shown in the figure below is pulled up to point 1 where it is released from rest. Take y = 31 mm .(Figure 1) Part a) Assuming no friction, calculate the speed at point 2. Part b) Calculate the speed at point 3. part c) Calculate the speed at point 4.

A) KE1+PE1=KE2+PE2 0+mgh₁=1/2mv²₂+0 AWS= 25 m/s B) KE1+PE1=KE2+PE2 1/2mv²₁+mgh₁=mgh₂+ 1/2mv²₂ AWS= 9.9 m/s C) KE1+PE1=KE2+PE2 0+mgh₁=mgh₂+ 1/2mv²₂ AWS 18 m/s

A 66-kgkg skier starts from rest at the top of a 1200-mm-long trail which drops a total of 180 mm from top to bottom. At the bottom, the skier is moving 11 m/s

How much energy was dissipated by friction? F = PE - KE F= mgh-.5mv^2 AWS: Efr = 110 k

A novice skier, starting from rest, slides down an icy frictionless 6.0 ∘∘ incline whose vertical height is 110 mm . How fast is she going when she reaches the bottom?

KE+PE=KE'+PE' KE- top KE'-bottom KE=0 PE'=0 0+mgh=1/2mv²+0 Solve for V AWS: 46 m/s

If the speed of a car is increased by 30%, by what factor will its minimum braking distance be increased, assuming all else is the same? Ignore the driver's reaction time. Express your answer to three significant figures.

KE₂= 1/2 mv² AWS: d1/d0 = 1.69

A 390 kgkg piano slides 2.7 mm down a 28 ∘incline and is kept from accelerating by a man who is pushing back on it parallel to the incline (see the figure(Figure 1)). Ignore friction. Part a: Determine the force exerted by the man. Express your answer to two significant figures and include the appropriate units. Part b: Determine the work done on the piano by the man. Express your answer to two significant figures and include the appropriate units. part c: Determine the work done on the piano by the force of gravity. Express your answer to two significant figures and include the appropriate units. part d: Determine the net work done on the piano. Express your answer to two significant figures and include the appropriate units

PART A: Fx-Ff=ma Aceleration is zero so Fsin (theta) - µmg cos(theta) = 0 F=µmg cos(theta)/ sign (theta) AWS: FP = 1800 N PART B: Wf= Ffdcos(180) AWS: WP = -4800 J PART C: Angle is 62 Wg = Fgdcos(62)= mgcos(62) AWS: Wg = 4800 J PART D: since the net force is 0 because the piano is no acceleraring, the net work done is also 0 wp+wg=0 AWS: Wnet = 0 J

A ski starts from rest and slides down a 28∘ incline 80 mm long. Part A) If the coefficient of friction is 0.090, what is the ski's speed at the base of the incline? Part B) If the snow is level at the foot of the incline and has the same coefficient of friction, how far will the ski travel along the level? Use energy methods.

Part A) Frictional force acting on incline = μ mg cosθ work done by friction = frictional force x displacement Potential energy of sli at height = mgh net energy at the base =potential- work done by friction This will be in the form of kinetic energy . 1/2 m v²= net energy AWS: V= 25 m/s PART B) Let the distance travelled be d work done by frictional force = μ mg x d Aws: L=350m

A 0.80-kg ball is thrown with a speed of 12 m/sm/s at an upward angle of 31 ∘. Use conservation of energy Part A: What is its speed at its highest point? Part B: How high does it go?

Part A: (1/2)mv^2=mgh Aws: Vtop= 10.3 m/s PART B: h=V^2/2g AWS: y=1.9m

A block of mass m is attached to the end of a spring (spring stiffness constant k, (Figure 1). The mass is given an initial displacement x0 from equilibrium, and an initial speed v0. Part a) Ignoring friction and the mass of the spring, use energy methods to find its maximum speed in terms of the given quantities. Part b) Find its maximum stretch from equilibrium in terms of the given quantities.

Part a) E=1/2mv²₀+1/2kx²₀ For the maximum speed of the block, the total mechanical energy of the system is kinetic energy as the potential energy is zero. Then, E=1/2mv² 1/2mv²=1/2mv²₀+1/2kx²₀ Solve for V vmax= √V0^2+k/mx0^2 Part b) For the maximum stretch of the block, the total mechanical energy of the system is the potential energy of the spring,asthe kinetic energy of the block is zero. Then, E=1/2mv²₀+1/2kx²₀ E=1/2kx²₀ 1/2kx²₀=1/2mv²₀+1/2kx²₀ solve for x AWS: √m/kv0^2+x0^2

A driver notices that her 1260-kgkg car, when in neutral, slows down from 95 km/hkm/h to 65 km/hkm/h in about 7.0 ss on a flat horizontal road. Part a) Approximately what power (watts) is needed to keep the car traveling at a constant 80 km/hkm/h? part B) Approximately what power (hphp) is needed to keep the car traveling at a constant 80 km/hkm/h?

Part a) KE = 2 mv2 AWS: P= 3.3*10^4 W Part b) KE is equal to net work (W) done Now power= W/t AWS: P= 45 hp

A 65.5-kgkg hiker starts at an elevation of 1240 mm and climbs to the top of a peak 2750 mm high. Part A) What is the hiker's change in potential energy? PART B) What is the minimum work required of the hiker? Part C) Can the actual work done be greater than this?

Part a) ∆PE= mgh₂-mgh₁ =mg(h₂-h₁) AWS: 9.69X10^5J Part b) equivalent to DeltaPE Part c) Yes, the actual work done can be greater

A 1300-NN crate rests on the floor. How much work is required to move it at constant speed 5.0 mm along the floor against a friction force of 210 NN . Express your answer to two significant figures and include the appropriate units. How much work is required to move it at constant speed 5.0 mm vertically. Express your answer to two significant figures and include the appropriate units.

Part a: w=FfdCos(theta) Force would be just friction because second law of motion the net force is equal to zero since moving w= 210(5)(1) AWS: Wp= 1.1 kj PART b: Work= force x distance force is weight 1200*5 AWS: WP = 6.5 kJ

When a flea (m = 500 μgg) is jumping up, it extends its legs 0.5 mmmm and reaches a speed of 0.50 m/sm/s in that time. How much work did the flea do during that time? Use g= 10 m/s2m/s2.

force exerted is equal to it's weight weight= mass x gavity work= force x distance AWS: 6.5×10'⁸J (to the negative 8)

Two bullets are fired at the same time with the same kinetic energy. If one bullet has twice the mass of the other, what is the ratio of the speed of the lighter bullet to the speed of the heavier? Which can do the most work?

kinetic energy= mv² 1/2*m_1*(V_1)^2 = 1/2*m_2*(V_2)^2 m_2 = 2*m_1 (Plug into above and solve for v1/v2 1/2*m_1*(V_1)^2 = m_1*(V_2)^2 V_1 = sqrt(2*(V_2)^2) V_1 = sqrt(2)*V_2 AWS: vlighter/vheavier = 1.41 Both bullets can do the same amount of work

Chris jumps off a bridge with a bungee cord (a heavy stretchable cord) tied around his ankle, (Figure 1). He falls for 15 mm before the bungee cord begins to stretch. Chris's mass is 80 kgkg and we assume the cord obeys Hooke's law, F=−kx, with 50 N/m If we neglect air resistance, estimate what distance d� below the bridge Chris's foot will be before coming to a stop. Ignore the mass of the cord (not realistic, however) and treat Chris as a particle. Express your answer to two significant figures and include the appropriate units.

mg(h+x)=.5kx^2 Aws: d= 57m

In the high jump, the kinetic energy of an athlete is transformed into gravitational potential energy without the aid of a pole. With what minimum speed must athlete leave the ground in order to lift his center of mass 2.05 mm and cross the bar with a speed of 0.50 m/sm/s ?

mgh₁+1/2mv²₁=mgh₂+1/2mv²₂ (factor out mass since constant) AWS: 4.6 m/s

If a car generates 18 hp when traveling at a steady 70 km/hkm, what must be the average force exerted on the car due to friction and air resistance?

power= force X velocity Convert to SI to Watts to m/s AWS: F= 690 N