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lord [1]
3 years ago
7

A person pushes horizontally with a force of 200. N on a 65.0 kg crate to move it across a level floor. The coefficient of kinet

ic friction is 0.270. (a) What is the magnitude of the frictional force? (b) What is the magnitude of the crate's acceleration? Use g=9.81 m/s2.
Physics
1 answer:
Alina [70]3 years ago
7 0

Answer:

a)f_k=172.17N

b)a_x=0.428m/s^2

Explanation:

The magnitude of the kinetic frictional force is given by the formula

f_k=\mu_k N

In this case, on the vertical direction the only forces are the weight and the normal force, and since the object does not move in that direction they must be equal. We have then:

f_k=\mu_k N=\mu_k W = \mu_k mg

Substituting values we get

f_k=\mu_k mg=(0.27)(65Kg)(9.81m/s^2)=172.17N

On the horizontal direction the only forces are the one the person exerts (F_p) and the kinetic friction opposing it, so the net force is:

F_x=F_p-f_k

We apply then Newton's 2nd Law to calculate the acceleration of the object in the horizontal direction:

ma_x=F_x=F_p-f_k

a_x=\frac{F_p-f_k}{m}

And substituting values we have

a_x=\frac{200N-172.17N}{65Kg}=0.428m/s^2

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