Question

A 200 kg crate is pulled along a level surface by an engine. the coefficient of friction between crate and surface is 0.4.

a. how much power must the engine deliver to move the crate at 5.0m/s.
b. how much work is done by the engine in 180s

Answer 1

Normal force = 200(9.81) = 1962 N
Friction force = 1962 * .4 = 784.8
power = force * distance / time = 784.8 * 5 = 3924 Watts

Watts = Joules / sec
3924 x 180 = 706,320 Joules

Answer 2

a) Power = P = 3924 watts

b) Work = W = 706320 J

Step-by-step explanation:

a) Mass of the block = m = 200 kg

Velocity = V = 5 m/s

Time = t = 180s

Coefficient of friction = µ = 0.4

Gravitational acceleration = g = 9.81 m/s^2

According to newton’s second law of motion:

F-f = 0

Where,

F = Normal force

f = Frictional force

so,

F = f = µmg = (0.4)(200)(9.81) = 784.8 N

We know that:

P = FV = (784)(5) = 3924 watts

b) We know that:

W = P × t = 3924 × 180 = 706320 J