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)

We know that work done by a force is calculated this way:

W = F.d

F=42.5N, d= 4.39 m

W= (42.5)(4.39)

W= 186.57J

Part (b)

The total heat generated is equal to the work done by the friction.

So:

-Wf = Ethermal1 + Ethermal2 (Eq.1)

Ethermal1 = the change in the thermal energy of the block

Ethermal2 = the change in the thermal energy of the floor

Fy = 0

===> N - mg = 0

====> N = m.g , f = .N

g = 9.78 (m/s2)

Wf = f.d = .N.d = .m.g.d

=0.637×4.11×9.8×-4.39

=-112.6J

Ethermal1= 30.2J

Substituting the value

Ethermal2 = 82.4 J

Part (c)

Work net= kinetic energy

Delta Kinetic energy = 186-112.6=73.4 J

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