Question

A boxed 15.0 kg computer monitor is dragged by friction 6.00 m up along the moving surface of a conveyor belt inclined at an angle of 36.3 ∘ above the horizontal. The monitor's speed is a constant 2.30 cm/s

How much work is done on the monitor by friction?

Express your answer in joules.

How much work is done on the monitor by gravity?

Express your answer in joules.

How much work is done on the monitor by the normal force of the conveyor belt?

Express your answer in joules.

Answer 1

here,

mass , m = 15 kg

s = 6 m

theta = 36.3 degree

constant speed , v = 2.3 cm/s

v = 0.023 m/s

as it is moving at constant speed

the net force is zero

the friction force = component of gravity

ff = m * g * sin(theta)

ff = 15 * 9.81 * sin(36.3) N = 87.11 N

the work done on the monitor by friction , Wff = ff * s

Wff = 87.11 * 6 J = 522.7 J

the work done on the monitor by gravity, Wg = - m * g * sin(theta) * s

Wg = - 87.11 * 6 J = - 522.7 J

the work done on the monitor by the normal force , Wn = N * s

Wn = 0 J