A horizontal force of magnitude 42.5 N pushes a block of mass 4.11 kg across a f

Question

A horizontal force of magnitude 42.5 N pushes a block of mass 4.11 kg across a floor where the coefficient of kinetic friction is 0.637. (a) How much work is done by that applied force on the block-floor system when the block slides through a displacement of 4.39 m across the floor? (b) During that displacement, the thermal energy of the block increases by 30.2 J. What is the increase in thermal energy of the floor? (c) What is the increase in the kinetic energy of the block?

A horizontal force of magnitude 42.5 N pushes a block of mass 4.11 kg across a floor where the coefficient of kinetic friction is 0.637. (a) How much work is done by that applied force on the block-floor system when the block slides through a displacement of 4.39 m across the floor? (b) During that displacement, the thermal energy of the block increases by 30.2 J. What is the increase in thermal energy of the floor? (c) What is the increase in the kinetic energy of the block?

Explanation / Answer

part a

work done by the applied force on system is = work done by applied force - work done by friction force

= 42.5*4.39 - 0..637*4.11*9.81*4.39 = 73.9254 J

part b

it is the loss by friction force so it will be same for the floor for and block beacuse by newton's third law every force have equal reaction so when force is same then work done is also equal and this will convert in heat to raise the temp.

part c

by energy coservation work done will store as kinetic energy

kinetic energy = work done by force on system = 73.924J by part a

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