Ch.7 #4 7,11,12,18 21,24 27

Transcription

1 Ch.7 #4 7,,,8, Picture the Problem: The farmhan pushes the hay horizontally. 88 N Strategy: Multiply the force by the istance because in this case the two point along the same irection. 3.9 m Solution: Apply equation 7 irectly: W 88 N 3.9 m 340 J 0.34 kj Insight: The 6 kg mass is unneee information unless we neee to know the amount of friction or the acceleration of the bale. 5. Picture the Problem: The chilren lift the bucket vertically. Strategy: Multiply the force by the istance because in this case the two point along the same irection. Solution: Apply equation 7 : W mg mg 4.70 m Now solve for m: W 0 J m g 9.8 m/s 4.70 m Insight: The applie force equals the weight as long as the bucket oes not accelerate kg 6. Picture the Problem: The pumpkin is lifte vertically then carrie horizontally. Strategy: Multiply the force by the istance because uring the lift the two point along the same irection. Solution:. (a) Apply equation 7 irectly: W mg 3. kg 9.8 m/s. m 38 J. (b) The force is perpenicular to the isplacement so W 0. Insight: You can still get tire carrying a pumpkin horizontally even though you re oing no work! 7. Picture the Problem: The suitcase is pushe horizontally. Strategy: Determine the applie force an solve for. W W 64 J Solution: Solve equation 7 for : f mg kg 9.8 m/s k k Insight: The applie force equals the friction force as long as the suitcase oes not accelerate m

2 . Picture the Problem: The wagon rolls horizontally but the force pulls upwar at an angle. Strategy: Use equation 7 3 keeping in min the angle between the force an the irection of motion. W cos 6 N 0.0 m cos 5 50 J 0.5 kj Solution: Use equation 7 3: o Insight: Only the component of the force along the irection of the motion oes any work. The vertical component of the force reuces the normal force a little.. Picture the Problem: The packing crate slies horizontally but the force pulls upwar at an angle. Strategy: Use equation 7 3 keeping in min the angle between the force an the irection of motion. Solution: Use equation 7 3: W cos 5 N 8.0 m cos J 0.67 kj Insight: Only the component of the rope force along the irection of the motion oes any work. The vertical component of the force reuces the normal force a little. riction must o 0.67 kj of work on the crate because the crate oes not gain or lose energy as it is pulle at constant spee. 8. Picture the Problem: A pitcher throws a ball at 90 mi/h an the catcher stops it in her glove. Strategy: Use the work energy theorem to answer the conceptual question. Solution:. (a) The ball gains kinetic energy from the action of the pitcher, so we conclue that the work one on the ball by the pitcher is positive.. (b) The ball loses kinetic energy from the action of the catcher, so we conclue that the work one on the ball by the catcher is negative. Insight: The answers are consistent with the previously escribe efinition of work: The pitcher oes positive work on the ball by exerting a force in the same irection as the motion of the ball. The catcher oes negative work on the ball because the force exerte by the catcher is opposite in irection to the motion of the ball. 9. Picture the Problem: The runner accelerates horizontally an runs in a straight line. Strategy: The work one equals the change in kinetic energy. Solution: in the change in kinetic energy: W K mv mv f i 73 kg 7.7 m/s 0 00 J. kj Insight: The runner s kinetic energy comes from the forces his muscles exert on his center of mass over the istance which his center of mass moves.

3 0. Picture the Problem: The fragment moves at high spee in a straight line. Strategy: Calculate the kinetic energy using equation 7 6. Solution: Apply equation 7 6 irectly: K mv 770 kg 0 m/s.7 0 J.7 MJ 7 Insight: The energy came from the work the rocket motor i in orer to place Skylab into orbit.. Picture the Problem: The bullet moves at high spee in a straight line. Strategy: Calculate the kinetic energy using equation 7 6, then use ratios to fin the new kinetic energies in (b) an (c). Solution:. (a) Apply equation 7 6 irectly:. (b) Use a ratio to preict the new kinetic energy: 3. (c) Use a ratio to preict the new kinetic energy: K mv kg 300 m/s 8030 J 8.03 kj new new ol ol ol ol new K 4 ol 4 K mv v so that K mv v 4 K 8.03 kj.0 kj new ol Knew mv v new ol 4 so that Kol mvol vol K 4K J 3. kj Insight: The kinetic energy is proportional to the velocity square, a useful thing to remember when calculating ratios. 4. Picture the Problem: The pine cone falls straight own uner the influence of gravity. Strategy: The work one by gravity equals the change in kinetic energy accoring to equation 7 7. The work one by gravity is always W = mgh as inicate in Example 7 an Conceptual Checkpoint 7. Solution:. (a) The work one by gravity on the pine cone equals the increase in its kinetic energy. Set the energies W K mgh mv v gh equal an solve for v: 9.8 m/s 6 m 8 m/s. (b) Air resistance i negative work because the spee an therefore the kinetic energy of the pine cone when it lane was reuce. Air resistance remove energy from the pine cone. Insight: Kinetic friction always oes negative work because the force is always opposite to the irection of motion. 5. Picture the Problem: The pine cone falls straight own for 6 m uner the influence of gravity. Strategy: The work one by air resistance is the ifference in kinetic energies between the air resistance an no air resistance cases. The work one by gravity is always W = mgh (see Example 7 an Conceptual Checkpoint 7 ). Solution:. (a) in the ifference in kinetic energies between the air resistance an no air resistance cases:. (b) The work one by air resistance equals the average force of air resistance times the istance the pine cone falls. It is negative because the upwar force is opposite to the ownwar istance travele. W K K K mv mgh f i f 0.40 kg 3 m/s 9.8 m/s 6 m 0 J W 0 J W so that 0.63 N upwar 6 m Insight: Kinetic friction always oes negative work because the force is always opposite to the irection of motion.

4 6. Picture the Problem: The object falls straight own uner the influence of gravity. Strategy: Use the epenence of kinetic energy upon mass an spee to answer parts (a) an (b). The work one by gravity can be foun from the change in the kinetic energy. Solution:. (a) Apply equation 7 6 irectly: K mv 0.40 kg 6.0 m/s 7. J 5 J. (b) Solve equation 7 6 for spee: K v m/s m 0.40 kg 3. (c) Calculate W K : W K Kf Ki 5 J 7. J 8 J Insight: As an object falls, the work one by gravity increases the kinetic energy of the object. 7. Picture the Problem: The runner slies horizontally on level groun over a istance of 3.40 m an comes to rest. Strategy: The work one by friction equals the negative of the kinetic energy the runner ha just before the slie. It also equals the force exerte by friction times the istance of the slie. Solution:. (a) Calculate W K : W K K K 0 mv W f i i 6.0 kg 4.35 m/s 587 J. (b) The work one by friction equals the average force of friction times the istance the player sli: W W kmg so that k mg 587 J 6.0 kg9.8 m/s 3.40 m k 0.84 Insight: Kinetic friction always oes negative work because the force is always opposite to the irection of motion. Ch.7 #35,4,44, Picture the Problem: The work one by the force is the area uner the force versus position graph. Strategy: The total work one by the force is the total area uner the graph from zero to 0.75 m. The work one from 0.5 m to 0.60 m is the area shae in gray in the figure at right. A the works one in each of the three segments to fin the total work. Solution:. (a) The total area uner the graph from 0 to 0.75 m:. (b) The area of the shae graph portion: Wtotal A 0.5 m N 0.45 J m0.6 N 0.5 m0.4 N m0.8 N W W 0.4 J Insight: The work is positive as long as the object moves from left to right (from small x to large x). Therefore the object gains energy as it moves from left to right.

5 4. Picture the Problem: orces of ifferent magnitues o ifferent amounts of work in ifferent intervals of time. Strategy: Use the efinition of power to etermine the ranking of the powers. Solution:. Calculate the powers prouce by each force: W 5 J. P 0.50 W t 0 s W3 6 J 4. P W t 8 s 3 W 3 J 3. P 0.60 W t 5 s W4 5 J 5. P4 0.0 W t 5 s 4 6. By comparing the magnitues of the powers we arrive at the ranking 4 3. Insight: If the amount of work an the time interval are each ouble, the power prouce remains the same. 44. Picture the Problem: This is a units conversion problem. Strategy: Use the conversion factors given in the insie front cover of the book. Solution: Convert the units: 6 W Pt kw 000 W/kW h 3600 s/h 3.60 J 3.6 MJ Insight: Electrical energy is sol in units of kwh, which are really units of energy in multiples of 3.6 million joules. 45. Picture the Problem: The fly oes work against gravity as it elevates its center of mass. Strategy: The power require is the force times the velocity, where the force is just the weight of the fly. Solution: Apply equation 7 3: P v mgv kg9.8 m/s 0.03 m/s P W 0.3 mw Insight: The energy an power require of the fly is higher than this because it isn t 00% efficient at converting foo energy into mechanical energy.