Work and Energy
A. 0.5 The KE of any object can be computed if the mass (m) and speed (v) are known. Simply use the equation KE=0.5*m*v2 In this case, the 10-N object has a mass of approximately 1 kg (use Fgrav = m*g). The speed is 1 m/s. Now plug and chug to yield KE of approximately 0.5 J.
A 10-Newton object moves to the left at 1 m/s. Its kinetic energy is approximately ____ Joules.
E. 180. m When a car skids to a stop, the work done by friction upon the car is equal to the change in kinetic energy of the car. Work is directly proportional to the displacement of the car (skidding distance) and the kinetic energy is directly related to the square of the speed (KE=0.5*m*v2). For this reason, the skidding distance is directly proportional to the square of the speed. So if the speed is tripled from 40 km/hr to 120 km/hr, then the stopping distance is increased by a factor of 9 (from 20 m to 9*20 m; or 180 m). A detailed discussion of the distance-speed squared relationship can be found at The Physics Classroom.
A 1000-kg car is moving at 40.0 km/hr when the driver slams on the brakes and skids to a stop (with locked brakes) over a distance of 20.0 meters. How far will the car skid with locked brakes if it is traveling at 120. km/hr?
C. twice as much The amount of work done by a force to displace an object is found from the equation W = F*d*cos(Theta) The force required to raise the car at constant speed is equivalent to the weight (m*g) of the car. Since the 2400-kg car weighs 2X as much as the 1200-kg car, it would require twice as much work to lift it the same distance.
A 1200 kg car and a 2400 kg car are lifted to the same height at a constant speed in a auto service station. Lifting the more massive car requires ____ work.
2.56 m/s vf = SQRT ( 2 • 9.8 m/s/s • 0.33 m) = SQRT(6.53 m/s) = 2.56 m/s
A 21.3-kg child positions himself on an inner-tube which is suspended by a 7.28-m long rope attached to a strong tree limb. The child and tube is drawn back until it makes a 17.4-degree angle with the vertical. The child is released and allowed to swing to and from. Assuming negligible friction, determine the child's speed at his lowest point in the trajectory.
46.1 m/s 0.5•(36.7 m/s)2 + (9.8 m/s/s)•(39.8 m) = 0.5 • vf2 673.4 m2/s2 + 390.0 m2/s2 = 0.5 • vf2 1063 m2/s2 = 0.5 • vf2 2127 m2/s2 = vf2 SQRT(2127 m2/s2) = vf vf = 46.1 m/s
A 221-gram ball is thrown at an angle of 17.9 degrees and a speed of 36.7 m/s from the top of a 39.8-m high cliff. Determine the impact speed of the ball when it strikes the ground. Assume negligible air resistance.
12.7 m/s 0.5•m•vi2 + m•g•hi + F•d•cos(theta) = 0.5•m•vf2 0.5•(29.1 kg)•(8.96 m/s)2 + (29.1 kg)•(9.8 m/s/s)•(8.21 m) + (30.3 N)(37.9 m)•cos(180 deg)= 0.5•(29.1 kg)•vf2 1168 J + 2341 J - 1148 J = (14.6 kg)•vf2 162.3 m2/s2 = vf2 vf = SQRT(162.3 m2/s2) = 12.7 m/s
A 29.1-kg sledder is traveling along a level area with a speed of 8.96 m/s when she approaches a gentle incline which makes an angle of 12.5 degrees with the horizontal. If the coefficient of friction between the sled and the incline is 0.109, then what will be her speed at the bottom of the inclined plane, located 8.21 m above the top of the incline
B. 10 The kinetic energy of the diver upon striking the water must be equal to the original potential energy. Thus, m*g*hi = KEf (50 kg)*(~10 m/s/s)*h = 5000 J So, h = ~10 m
A 50-kg platform diver hits the water below with a kinetic energy of 5000 Joules. The height (relative to the water) from which the diver dove was approximately ____ meters.
C. the same The power is the rate at which work is done (or energy is used). Power is found by dividing work by time. It requires the same amount of work to do these two jobs (see question #23) and the same amount of time. Thus, the power is the same for both tasks.
A 50.0 kg crate is lifted to a height of 2.0 meters in the same time as a 25.0 kg crate is lifted to a height of 4 meters. The rate at which energy is used (i.e., power) in raising the 50.0 kg crate is ____ as the rate at which energy is used to lift the 25.0 kg crate.
a. Delta KE= 0 J b. Delta PE= +21900 J c. W= +21900 J d. P= 57.0 Watts a. The speed of the hiker is constant so there is no change in kinetic energy - 0 J. b. The potential energy change can be found by subtracting the initial PE (0 J) from the final PE (m*g*hf). The final potential energy is 21888 J [from (51.7 kg)*(9.8 m/s/s)*(43.2 m)] and the initial potential energy is 0 J. So Delta PE = +21900 J (rounded from 21888 J). c. The work done upon the hiker can be found using the work-energy theorem. The equation reduces to Wnc = PEf (PEi = 0 J since the hiker starts on the ground; and KEi = KEf since the speed is constant; these two terms can be dropped from the equation since they are equal). The final potential energy is 21888 J [from (51.7 kg)*(9.8 m/s/s)*(43.2 m)]. So W = +21900 J (rounded from 21888 J). d. The power of the hiker can be found by dividing the work by the time. P=W/t=(21888 J)/(384 s) = 57.0 Watts
A 51.7-kg hiker ascends a 43.2-meter high hill at a constant speed of 1.20 m/s. If it takes 384 s to climb the hill, then determine ... a. kinetic energy change of the hiker. b. the potential energy change of the hiker. c. the work done upon the hiker. d. the power delivered by the hiker.
116 m KEi + PEi + Wnc = 0 KEi + PEi + Wincline + Wlevel = 0 0.5•(62.9 kg)•(12.9 m/s)2 + (62.9 kg)•(9.8 m/s)•(123 m) + (72.3 N)•(505 m)•cos(180 deg) + (384 N)•(x)•cos(180 deg) = 0 J 5233 J + 75820 J - 36512 J - 384 x = 0 J 44541 J = 384 x 116 m = x
A 62.9-kg downhill skier is moving with a speed of 12.9 m/s as he starts his descent from a level plateau at 123-m height to the ground below. The slope has an angle of 14.1 degrees and a coefficient of friction of 0.121. The skier coasts the entire descent without using his poles; upon reaching the bottom he continues to coast to a stop; the coefficient of friction along the level surface is 0.623. How far will he coast along the level area at the bottom of the slope
A. less than 100. During any given motion, if non-conservative forces do work upon the object, then the total mechanical energy will be changed. If non-conservative forces do negative work (i.e., Fnc*d*cos(Theta) is a negative number), then the final TME is less than the initial TME. In this case, air resistance does negative work to remove energy from the system. Thus, when the ball returns to its original height, their is less TME than immediately after it was thrown. At this same starting height, the PE is the same as before. The reduction in TME is made up for by the fact that the kinetic energy has been reduced; the final KE is less than the initial KE.
A ball is projected into the air with 100 J of kinetic energy. The kinetic energy is transformed into gravitational potential energy on the path towards the peak of its trajectory. When the ball returns to its original height, its kinetic energy is ____ Joules. Do consider the effects of air resistance
A. zero J For any given situation, the work done by a force can be calculated using the equation W = F*d*cos(Theta) where F is the force doing the work, d is the displacement of the object, and Theta is the angle between the force and the displacement. In this specific situation, the child is applying an upward force on the box (he is carrying it) and the displacement of the box is horizontal. The angle between the force (vertical) and the displacement (upward) vectors is 90 degrees. Since the cosine of 90-degrees is 0, the child does not do any work upon the box. A detailed discussion of a similar situation (the waiter and the tray of food) can be found at The Physics Classroom.
A child lifts a box up from the floor. The child then carries the box with constant speed to the other side of the room and puts the box down. How much work does he do on the box while walking across the floor at constant speed?
D. work, power Power refers to the rate at which work is done. Thus, doing two jobs - one slowly and one quickly - involves doing the same job (i.e., the same work and same force) at different rates or with different power.
A job is done slowly, and an identical job is done quickly. Both jobs require the same amount of ____, but different amounts of ____. Pick the two words which fill in the blanks in their respective order.
D. 5000 The use of the work-energy theorem and a simple analysis will yield the solution to this problem. Initially, there is only PE; finally, there is only KE. Assuming negligible air resistance, the kinetic energy of the diver upon hitting the water is equal to the potential energy of the diver on top of the board. PEi = KEf m*g*hi = KEf Substituting 500 N for m*g (500 N is the weight of the diver, not the mass) and 10 m for h will yield the answer of 5000 J.
A platform diver weighs 500 N. She steps off a diving board that is elevated to a height of 10 meters above the water. The diver will possess ___ Joules of kinetic energy when she hits the water.
13.2 J There are two methods of solving this problem. The first method involves using the equation W = F*d*cos(Theta) where F=20.8 N, d=0.636 m, and Theta=0 degrees. (The angle theta represents the angle betwee the force and the displacement vector; since the force is applied parallel to the incline, the angle is zero.) Substituting and solving yields W = F*d*cos(Theta) = (20.8 N)*(0.636 m)*cos(0) = 13.2 J. The second method is to recognize that the work done in pulling the cart along the incline at constant speed changes the potential energy of the cart. The work done equals the potential energy change. Thus, W=Delta PE = m*g*(delta h) = (3.00 kg)*(9.8 m/s/s)*(0.45 m) = 13.2 J
A student applies a force to a cart to pull it up an inclined plane at a constant speed during a physics lab. A force of 20.8 N is applied parallel to the incline to lift a 3.00-kg loaded cart to a height of 0.450 m along an incline which is 0.636-m long. Determine the work done upon the cart and the subsequent potential energy change of the cart
A. A rightward applied force is used to displace a television set to the right. - Answer: A - 0 degrees B. The force of friction acts upon a rightward-moving car to bring it to a stop. - Answer: B - 180 degrees C. A waiter uses an applied force to balance the weight of a tray of plates as he carries the tray across the room. - Answer: C - 90 degrees D. The force of air resistance acts upon a vertically-falling skydiver. - Answer: B - 180 degrees E. The force of friction acts upon a baseball player as he slides into third base. - Answer: B - 180 degrees F. An applied force is used by a freshman to lift a World Civilization book to the top shelf of his locker. - Answer: A - 0 degrees G. A bucket of water is tied to a string and tension supplies the centripetal force to keep it moving in a circle at constant speed. - Answer: C - 90 degrees H. An applied force acting at 30-degrees to the horizontal is used to displace an object to the right. - Answer: D - 30 degrees I. A group of football players use an applied force to push a sled across the grass. - Answer: A - 0 degrees J. The tension in the elevator cable causes the elevator to rise at a constant speed. - Answer: A - 0 degrees K. In a physics lab, an applied force is exerted parallel to a plane inclined at 30-degrees in order to displace a cart up the incline. - Answer: A - 0 degrees L. An applied force is exerted upwards and rightwards at an angle of 30-degrees to the vertical in order to displace an object to the right. - Answer: E - 60 degrees M. A child rests on the seat of a swing which is supported by the tension in its cables; he swings from the highest position to its lowest position. - Answer: C - 90 degrees
Consider the following physical situations. For each case, determine the angle between the indicated force (in boldface type) and the displacement ("theta" in the work equation). a. 0 degrees b. 180 degrees c. 90 degrees d. 30 degrees e. 60 degrees
A. A cable is attached to a bucket and the force of tension is used to pull the bucket out of a well. -positive work B. Rusty Nales uses a hammer to exert an applied force upon a stubborn nail to drive it into the wall. - positive work C. Near the end of the Shockwave ride, a braking system exerts an applied force upon the coaster car to bring it to a stop. - negative work D. The force of friction acts upon a baseball player as he slides into third base. - negative work E. A busy spider hangs motionless from a silk thread, supported by the tension in the thread. - no work F. In baseball, the catcher exerts an abrupt applied force upon the ball to stop it in the catcher's mitt. - negative work G. In a physics lab, an applied force is exerted parallel to a plane inclined at 30-degrees in order to displace a cart up the incline. - positive work H. A pendulum bob swings from its highest position to its lowest position under the influence of the force of gravity. - positive work
Consider the following physical situations. Identify whether the indicated force (in boldface type) does positive work, negative work or no work.
P= 849 Watts Eddy's power is found by dividing the work which he does by the time in which he does it. The work done in elevating his 65.0-kg mass up the stairs is determined using the equation W = F*d*cos(Theta) where F = m*g = 637 N (the weight of the 65.0 kg object), d =1.60 m and Theta = 0 degrees (the angle between the upward force and the upward displacement). Solving for W yields 1019.2 Joules. Now divide the work by the time to determine the power: P = W/t = (1019.2 J)/(1.20 s) = 849 Watts
Eddy, whose mass is 65.0-kg, climbs up the 1.60-meter high stairs in 1.20 s. Approximate Eddy's power rating
A. A force is regarded as a conservative force if it does work but does not remove mechanical energy from a system of objects. (sort of) C. The force of gravity and elastic (spring) force are both examples of a conservative forces. D. Applied forces, air resistance, friction forces, and tension are common examples of non-conservative forces. G. If the only forces which do work upon an object are conservative forces, then the object will conserve its mechanical energy. H. If the sum of an object's KE and PE is remaining constant, then non-conservative forces are NOT doing work. I. If work is NOT done on an object by a non-conservative force, then the object will experience a transformation of energy from kinetic to potential energy (or vice versa) (sort of) J. An object starts from an elevated position with 50 J of potential energy and begins its fall towards the ground. If non-conservative forces can be assumed to NOT do work, then at some point during the fall the object will have 20 J of potential energy and 30 J of kinetic energy.
Which of the following statements are true about conservative and non-conservative forces? Include all that apply.
A. The total amount of mechanical energy of an object is the sum of its potential energy and the kinetic energy. E. A bowling ball is mounted from a ceiling by way of a strong cable. It is drawn back and released, allowed to swing as a pendulum. As it swings from its highest position to its lowest position, the total mechanical energy is mostly conserved. F. When a friction force does work on an object , the total mechanical energy of that object is changed. G. The total mechanical energy of an object remains constant if the only forces doing work on the object are conservative forces. H. If an object gains mechanical energy, then one can be certain that a non-conservative force is doing work.
Which of the following statements are true about mechanical energy? Include all that apply.
B. Potential energy is the energy stored in an object due to its position. D. The gravitational potential energy of an object is dependent upon the mass of the object. E. If the mass of an elevated object is doubled, then its gravitational potential energy will be doubled as well. F. Gravitational potential energy is lost as objects free-fall to the ground. G. The higher that an object is, the more potential energy which it will have. H. The unit of measurement for potential energy is the Joule.
Which of the following statements are true about potential energy? Include all that apply.
A. Power is a time-based quantity. B. Power refers to how fast work is done upon an object. D. A force is exerted on an object to move it at a constant speed. The power delivered by this force is the magnitude of the force multiplied by the speed of the object. E. The standard metric unit of power is the Watt. I. A 300-Newton force is applied to a skier to drag her up a ski hill at a constant speed of 1.5 m/s. The power delivered by the toe rope is 450 Watts.
Which of the following statements are true about power? Include all that apply.
A. Work is a form of energy. C. Units of work would be equivalent to a Newton times a meter. D. A kg•m2/s2 would be a unit of work. H. A force is applied by a chain to a roller coaster car to carry it up the hill of the first drop of the Shockwave ride. This is an example of work being done. I. The force of friction acts upon a softball player as she makes a headfirst dive into third base. This is an example of work being done. K. A force acts upon an object to push the object along a surface at constant speed. By itself, this force must NOT be doing any work upon the object. N. An object is moving to the right. A force acts leftward upon it. This force is doing negative work. O. A non-conservative force is doing work on an object; it is the only force doing work. Therefore, the object will either gain or lose mechanical energy.
Which of the following statements are true about work? Include all that apply.
C. both require the same amount of work Work involves a force acting upon an object to cause a displacement. The amount of work done is found by multiplying F*d*cos(Theta). The equation can be used for these two motions to find the work. Lifting a 50 kg crate vertically 2 meters W = (~500 N)*(2 m)*cos(0) W = ~1000 N (Note: The weight of a 50-kg object is approximately 500 N; it takes 500 N to lift the object up.) Lifting a 25 kg crate vertically 4 meters W = (~250 N)*(4 m)*cos(0) W = ~1000 N (Note: The weight of a 25-kg object is approximately 250 N; it takes 250 N to lift the object up.)
Which requires more work: lifting a 50.0 kg crate a vertical distance of 2.0 meters or lifting a 25.0 kg crate a vertical distance of 4.0 meters?
514 N Change in KE = KEf - KEi = 0 J - 0.5 • (0.163 kg) • (39.8 m/s)2 = -129 J The force can be determined by setting this value equal to the work and using the expression for work into the equation: Wnc = -129 J F • d • cos(theta) = -129 J F • (0.251 m) • cos(180 deg) = -129 J F = 514 N
A baseball player catches a 163-gram baseball which is moving horizontally at a speed of 39.8 m/s. Determine the force which she must apply to the baseball if her mitt recoils a horizontal distance of 25.1 cm.
a. W= -324000 J b. KEi = +324000 J c. a= -8.16 m/s/s d. vi= 27.2 m/s a. The work done upon the car can be found using the equation W = F*d*cos(Theta) where F=7160 N, d=45.2 m, and Theta=180 degrees (the force is in the opposite direction as the displacement). Substituting and solving yields -323632 J (rounded to -324000 J). b. The initial kinetic energy can be found using the work-energy theorem. The equation reduces to KEi + Wnc = 0 (PEi and PEf = 0 J since the car is on the ground; and KEf = 0 J since the car is finally stopped). Rearrange the equation and it takes the form KEi = -Wnc . So KEi = +324000 J (rounded from +323632 J). c. The acceleration of the car can be found using Newton's second law of motion: Fnet = m*a The friction force is the net force (since the up and down forces balance) and the mass is 878 kg. Substituting and solving yields a = -8.16 m/s/s. d. The initial velocity of the car can be found using the KE equation: KE = 0.5*m*v2 where m=878 kg and KEi=323632 J. Substituting and solving for velocity (v) yields v = 27.2 m/s. (A kinematic equation could be also used to find the initial velocity.)
An 878-kg car skids to a stop across a horizontal surface over a distance of 45.2 m. The average force acting upon the car is 7160 N. Determine .. a. the work done upon the car. b. the initial kinetic energy of the car. c. the acceleration of the car. d. the initial velocity of the car.
A. 50 A drawn arrow has 50 J of stored energy due to the stretch of the bow and string. When released, this energy is converted into kinetic energy such that the arrow will have 50 J of kinetic energy upon being fired. Of course, this assumes no energy is lost to air resistance, friction or any other non-conservative forces and that the arrow is shot horizontally
An arrow is drawn back so that 50 Joules of potential energy is stored in the stretched bow and string. When released, the arrow will have a kinetic energy of ____ Joules.
D. energy An object at rest absolutely cannot have speed or velocity or acceleration. However, an object at rest could have energy if there is energy stored due to its position; for example, there could be gravitational or elastic potential energy.
An object at rest may have __________.
~150 J The work done upon an object is found with the equation W = F*d*cos(Theta) In this case, the d=6.0 m; the F=24.5 N (it takes 24.5 N of force to lift a 2.5-kg object; that's the weight of the object), and the angle between F and d (Theta) is 0 degrees. Substituting these values into the above equation yields W = F*d*cos(Theta) = (24.5 N)*(6 m)*cos(0) = ~150 J (147 J)
Approximate the work required lift a 2.5-kg object to a height of 6.0 meters. PSYW
C. 250,000 The use of the work-energy theorem and a simple analysis will yield the solution to this problem. Initially, there is only PE; finally, there is neither PE nor KE; non-conservative work has been done by an applied force upon the falling object. The work-energy equation can be written as follows. PEi + Wnc = 0 PEi = - Wnc m*g*hi = - F*d*cos(Theta) Substituting 2500 N for m*g (2500 N is the weight of the driver, not the mass); 10.0 m for h; 0.100 m for the displacement of the falling object as caused by the upward applied force exerted by the post; and 90 degrees for Theta (the angle between the applied force and the displacement of the falling object) will yield the answer of 250000 N for F.
During a construction project, a 2500 N object is lifted high above the ground. It is released and falls 10.0 meters and drives a post 0.100 m into the ground. The average impact force on the object is ____ Newtons.
B. 100 Quite surprisingly to many, each ball would hit the ground with the same speed. In each case, the PE+KE of the balls immediately after being thrown is the same (they are thrown with the same speed from the same height). Upon hitting the ground, they must also have the same PE+KE. Since the PE is zero (on the ground) for each ball, it stands to reason that their KE is also the same. That's a little physics and a lot of logic - and try not to avoid the logic part by trying to memorize the answer.
Luke Autbeloe stands on the edge of a roof throws a ball downward. It strikes the ground with 100 J of kinetic energy. Luke now throws another identical ball upward with the same initial speed, and this too falls to the ground. Neglecting air resistance, the second ball hits the ground with a kinetic energy of ____ J.
0.170 m W = Change in KE F • d • cos(theta) = -1.80 J (10.6 N) • (d) • cos(180 deg) = -1.80 J - (10.6 N) • (d) = -1.80 J d = 0.170 m
Pete Zaria applies a 11.9-Newton force to a 1.49-kg mug of root beer in order to accelerate it from rest over a distance of 1.42-m. Once released, how far will the mug slide along the counter top if the coefficient of friction is 0.728?
F= 4.86*10^3 N The work energy theorem can be written as KEi + PEi + Wnc = KEf + PEf The PEi and PEf can be dropped from the equation since they are both 0 (the height of the car is 0 m). The KEf can also be dropped for the same reason (the car is finally stopped). The equation simplifies to KEi + Wnc = 0 The expressions for KE (0.5*m*v2) and Wnc (F*d*cos[Theta]) can be substituted into the equation: 0.5*m*vi2 + F*d*cos[Theta] = 0 where m=988 kg, vi=21.2 m/s, d=45.7 m, and Theta = 180 degrees. Substituting and solving for F yields 4.86*103 N.
Use the work-energy theorem to determine the force required to stop a 988-kg car moving at a speed of 21.2 m/s if there is a distance of 45.7 m in which to stop it.
B. 50 W This is a relatively simple plug-and-chug into the equation P=W/t with W=1000. J and t=20.0 s.
Using 1000. J of work, a small object is lifted from the ground floor to the third floor of a tall building in 20.0 seconds. What power was required in this task?
B. If an object is at rest, then it does not have any kinetic energy. G. Kinetic energy is a scalar quantity. H. An object has a kinetic energy of 40 J. If its mass were twice as much, then its kinetic energy would be 80 J. K. An object can never have a negative kinetic energy.
Which of the following statements are true about kinetic energy? Include all that apply.
