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.

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)

onyebuchinnaji onyebuchinnaji · Answer 2 · 2021-11-01T12:47:03+01:00

The magnitude of the average frictional force that acts on the box is [tex]( \frac{mv_0^2}{2 x_1})[/tex].

The given parameters;

initial speed of the box, = v₀
distance traveled by the box = x₁

Apply the principle of work-energy theorem, the work done by the frictional force is equal to the kinetic energy of the box.

[tex]\mu_k (F_k\times x_1)= \frac{1}{2} mv_0^2 \\\\F_k = \frac{mv_0^2}{2 x_1 \mu_k}[/tex]

where;

[tex]F_k[/tex] is the frictional force
[tex]\mu_k[/tex] is the coefficient of friction

Since it was advised to exclude the coefficient of friction, the frictional force in terms of m, v₀ and x₁ is expressed as follows;

[tex]F_k = \frac{mv_0^2}{2 x_1}[/tex]

Thus, the magnitude of the average frictional force that acts on the box is[tex]( \frac{mv_0^2}{2 x_1})[/tex]

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