Question

A factory worker pushes a 30.0-kg crate a distance of 4.50 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and the floor is 0.250.
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(a) What magnitude of force must the worker apply? (b) How much work is done on the crate by this force? (c) How much work is done on the crate by friction? (d) How much work is done by the normal force? By gravity? (e) What is the total work done

Answer 1

here,

the mass of crate , m = 30 kg

s = 4.5 m

coefficient of kinetic friction , uk = 0.25

a)

as it is moving at constant velocity

the acceleration of crate is zero

the horizontal force applied by the worker , F = uk * m * g

F = 0.25 * 30 * 9.81 N = 73.58 N

b)

the work done on the crate by this force , W1 = F * s

W1 = 73.58 * 4.5 J = 330.1 J

c)

the work done by friction force , W2 = - ff * s

W2 = - uk * m * g * s

W2 = - 0.25 * 30 * 9.81 * 4.5 J

W2 = - 330.1 J

d)

the work done by gravity , Wg = m * g * s* cos(90) = 0 J

the work done by normal force , Wn = N * s * cos(90) = 0 J

e)

the total work done , Wt = W1 + W2 + Wg+ Wn = 0 J