A 37.5 kg box initially at rest is pushed 4.05 m along a rough, horizontal floor with a constant applied horizontal force of 150 N. If the coefficient of friction between box and floor is 0.300, find the following. (a) the work done by the applied force

J

(b) the increase in internal energy in the box-floor system due to friction

J

(c) the work done by the normal force

J

(d) the work done by the gravitational force

J

(e) the change in kinetic energy of the box

J

(f) the final speed of the box

m/s

a) 607.5 J

b) 160.531875 J

c) 0 J

d) 0 J

e) 2.925 m/s

Explanation:

The given data : -

Mass of the box (m) = 37.5 kg. Displacement made by box (x) = 4.05 m. Horizontal force (F) = 150 N. The co-efficient of friction between box and floor (μ) = 0.3 Gravitational force (N) = m * g = 37.5 * 9.81 = 367.875

Solution:-

a) The work done by applied force (W)

W = force applied * displacement = 150 * 4.05 = 607.5 J

b) The increase in internal energy in the box-floor system due to friction.

Frictional force (f) = μ * N = 0.3 * 367.875 = 110.3625 N

Change in internal energy = change in kinetic energy.

ΔU = (K. E) ₂ - (K. E) ₁

Since the initial velocity is zero so the (K. E) ₁ = 0

ΔU = (K. E) ₂ = (F - f) * (x) = (150 - 110.3625) * 4.05 = 160.531875 J

c) The work done by the normal force.

Displacement of box vertically = 0

W = force applied * displacement = 367.875 * 0 = 0 J

d) The work done by the gravitational force.

Displacement of box vertically = 0

W = force applied * displacement = 367.875 * 0 = 0 J

e) The change in kinetic energy of the box

(K. E) ₂ - (K. E) ₁ = (K. E) ₂ - 0 = (F - f) * (x) = (150 - 110.3625) * 4.05 = 160.531875 J

f) The final speed of the box

(K. E) ₂ = 160.531875 J = 0.5 * 37.5 * v²

v² = 8.56

v = 2.925 m/s.