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

friction force acting on block = 4.11*10*0.637 = 26.187 N

force applied on the block = 42.5N

a) work done by applied force = 42.5*4.39 = 186.575 Joule

b) increase in thermal energy of the floor is same as of box hence 30.2 J

c) increse in kinetic energy of box = 186.575 - 26.187*4.39 = 71.61 J

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