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A box slides on a level floor. It is slowing with a constant

acceleration of magnitude 2.0 m/s2
. What is the coefficient of
kinetic friction between the box and the floor?

User Tahagh
by
7.2k points

1 Answer

1 vote

Answer:

Step-by-step explanation:

We can use the following equations of motion to solve the problem:

v^2 = u^2 + 2as

where v is the final velocity, u is the initial velocity, a is the acceleration, s is the distance travelled.

In this case, the box is slowing down, so the initial velocity u is greater than the final velocity v. We can use a negative sign to indicate that the acceleration is opposite to the initial velocity.

Let us assume that the mass of the box is m and the coefficient of kinetic friction is μ. The force of friction acting on the box is given by f = μmg, where g is the acceleration due to gravity.

Since the acceleration of the box is 2.0 m/s^2, we have

f = ma

μmg = m(-2.0)

μg = -2.0

μ = -2.0/g

Substituting g = 9.8 m/s^2, we get

μ = -0.204

Since the coefficient of friction cannot be negative, we take the absolute value and obtain:

μ = 0.204

Therefore, the coefficient of kinetic friction between the box and the floor is approximately 0.204.

User Amgad Fahmi
by
7.5k points