9. You push exactly horizontally with 770 N on a 150-kg crate, sliding it along
ID: 2039535 • Letter: 9
Question
9. You push exactly horizontally with 770 N on a 150-kg crate, sliding it along a level floor with friction. The crate starts from rest and moves 12 m in the direction you are pushing, over a timespan of 30. s. a. What is your average power output during the 30. s? B. 210 W C. 310 W D. 410 W E. 510 W b. If the crate's final speed is 1.7 m/s, what is the power dissipated by friction? A. 66 W B. 140 W C. 190 W D. 240 W E. 300 W c. Later, suppose that you encounter a patch of floor that has just the right amount of kinetic friction, so that the crate moves at a constant speed as you continue to exert your 750-N horizontal force. During this time, how does the power dissipated by the friction compare to the power produced by you?Explanation / Answer
(a) work done by me = Force×displacement
= 770×12 = 9240 joule
Power dissipated = work done/taken time
= 9240/30 = 308 watt option (c) 310 watt
(B) Kinetic energy of block = 1/2mv2
= 0.5×150×(1.7)2
= 216.75 joule
Energy lost to friction = work done by me - kinetic energy of block
= 9240 - 216.75 = 9023.25 joule
Power dissipated by friction = 9023.25/30 = 300.775 watt
= 300 watt
(c)
If we move the block with constant speed then there is no change in kinetic energy of the block . So whole power will be dissipated into friction power.
In this case power dissipated by friction force = 310 watt
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