Question

A worker pushed a 36.0 kg block 6.60 m along a level floor at constant speed with a force directed 23.0° below the horizontal. If the coefficient of kinetic friction between block and floor was 0.400, what were (a) the work done by the worker's force and (b) the increase in thermal energy of the block-floor system?

Answer 1

Since given that speed is constant, So acceleration is zero, whoch means net force in horizontal direction will be zero.

Fnet = Fapplied - Ff = 0

Fa = Ff

F*cos A = uk*N

N = W - F*sin A = mg - F*sin A

F*cos A = uk*m*g + uk*F*sin A

F = uk*m*g/(cos A - uk*sin A)

F = 0.4*36*9.8/(cos 23 deg - 0.4*sin 23 deg) = 184.66 N

Now

Workdone by Worker will be

W = F*d*cos A

W = 184.66*6.60*cos 23 deg

W = 1121.9 J

Part B.

Since we know that

Ff = uk*(mg + F*sin A)

Increase in thermal energy for the system will be

dE = Ff*d

dE = 0.4*(36*9.8 + 184.66*sin 23 deg)*6.6

dE = 1121.9 J

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