Question

2. A factory worker pushes a 40.0 kg crate a distance of 4.5 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 0.25. (a) What magnitude of force must the worker apply? (1 mark) (b) Write the force in part (a) as a row vector (1/2 mark) (c) Write the displacement as a row vector (1/2 mark) product of two row vectors (1 mark) (e) How much work is done on the crate by friction? (1 mark) (f) How much work is done on the crate by the normal force? By gravity? (1 mark) (g) What is the total work done on the crate? (1 mark)

Answer 1

(a) Mass of the crate, m = 40.0 kg

Distance, d = 4.5 m

Coefficient of kinetic friction, = 0.25

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Since the worker pushes the crate with constant velocity.

So, applied force, F = Frictional force acts on the crate

=> F = *m*g = 0.25*40.0*9.8 = 98.0 N

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(b) Write the force in unit vector notations considering force is applied in the right direction and displacement of the crate is in the right direction.

So, vector F = 98.0i N

(c) Displacement vector, d = 4.5i m

(d) Work done by the factory worker, W = (vector F) . (vector d)

= (98.0i) . (4.5i) = 441.0 J

(e) Work done on the crate by friction, Ff = -W = -441.0 J

(f) Since normal force is at 90 deg to the displacement of the crate.

So, Work done by the normal force = 0

Like above -

work done by the gravity = 0

(g) Total work done on the crate = 441.0 - 441.0 = 0 J