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**

**Please Upvote.**

**Comment below if you have any doubt.**

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