Question

A horizontal force of magnitude 46.3 n pushes a block of mass 4.14 kg across a floor where the coefficient of kinetic friction is 0.609. (a) how much work is done by that applied force on the block-floor system when the block slides through a displacement of 4.25 m across the floor? (b) during that displacement, the thermal energy of the block increases by 38.2 j. what is the increase in thermal energy of the floor? (c) what is the increase in the kinetic energy of the block?

Answer 1

A) Calling F the intensity of the horizontal force and d the displacement of the block across the floor, the work done by the horizontal force is equal to
$W=Fd = (46.3 N)(4.25 m)=196.8 J$

b) The work done by the frictional force against the motion of the block is equal to:
$W_f = -F_f d =- (\mu mg) d =-(0.609)(4.14 kg)(9.81 m/s^2)(4.25 m)=$
$=-105.1 J$
Part of these 105.1 Joules of work becomes increase of thermal energy of the block ($\Delta E_B$), and part of it becomes increase of thermal energy of the floor ($\Delta E_F$). We already know the increase in thermal energy of the block (38.2 J), so we can find the increase in thermal energy of the floor:
$\Delta E_F = 105.1 J - \Delta E_B = 105.1 J-38.2 J=66.9 J$

c) The net work done on the block is the work done by the horizontal force F minus the work done by the frictional force (the frictional force acts against the motion, so we must take it with a negative sign):
$W_{net}=W-W_f=196.8 J-105.1 J=91.7 J$
For the work-energy theorem, the work done on the block is equal to its increase of kinetic energy:
$W_{net} = \Delta K$
So, we have $\Delta K=+91.7 J$