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Xelga [282]
3 years ago
13

The box leaves position x=0 with speed v0. The box is slowed by a constant frictional force until it comes to rest at position x

=x1. Find Ff, the magnitude of the average frictional force that acts on the box. (Since you don't know the coefficient of friction, don't include it in your answer.)
Physics
1 answer:
Diano4ka-milaya [45]3 years ago
4 0

Answer:

Ff=m\times \dfrac{V_o^2}{2X_1}

Explanation:

Given that

At X=0 V=Vo

At X=X1  V=0

As we know that friction force is always try to oppose the motion of an object. It means that it provide acceleration in the negative direction.

We know that

V^2=U^2-2aS

0=V_o^2-2a X_1

a=\dfrac{V_o^2}{2X_1}

So the friction force on the box

Ff= m x a

Ff=m\times \dfrac{V_o^2}{2X_1}

Where m is the mass of the box.

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