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3 years ago

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

2 answers:

4 0

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

N76 [4]3 years ago

3 0

Answer:

a) Power = P = 3924 watts

b) Work = W = 706320 J

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

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The question is incomplete. Here is the complete question.

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