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.) Express the frictional force in terms of m, v0, and x1.

Question

Physics

Tanner

18 October, 08:09

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.) Express the frictional force in terms of m, v0, and x1.

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Emerson · Answer 1

a = Vo² / (2X₁)

Fr = mVo² / (2X₁)

Explanation:

Given that

Initial velocity = Vo

As we know that friction always tried to oppose the motion of the object. That is why acceleration due to friction acts opposite to the motion of the object. Lets take acceleration is a.

We also know that

v²=u² + 2 a s

v=final speed

u=initial speed

a = acceleration

s = distance

Here given that final speed is zero.

So

0²=Vo² - 2 x a x X₁ (negative sign because acceleration in the opposite to the displacement)

a = Vo² / (2X₁)

So the average friction force Fr

Fr = m a

Fr = mVo² / (2X₁)