Question

A horizontal force of 750 N is needed to overcome the force of static friction between a level floor and a 250-kg crate. What is the acceleration of the crate if the 750-N force is maintained after the crate begins to move and the coefficient of kinetic friction is 0.16?

Answer 1

First, we need to find the force of static friction that was initially acting on the crate to keep it from moving:

\( f_s = \mu_s \cdot N \),
where:
\( f_s \) = force of static friction
\( \mu_s \) = coefficient of static friction
\( N \) = normal force

The normal force is equal to the weight of the crate, which can be found using:
\( N = m \cdot g \),
where:
\( m \) = mass of the crate
\( g \) = acceleration due to gravity (9.8 m/s^2)

\( N = 250 \, kg \cdot 9.8 \, m/s^2 = 2450 \, N \)

Now we can calculate the force of static friction:
\( f_s = 0.16 \cdot 2450 \, N = 392 \, N \)

Since the crate is now moving, the force of kinetic friction comes into play:
\( f_k = \mu_k \cdot N \),
where:
\( f_k \) = force of kinetic friction
\( \mu_k \) = coefficient of kinetic friction

\( f_k = 0.16 \cdot 2450 \, N = 392 \, N \)

Since the applied force is greater than the force of kinetic friction:
\( F_{\text{net}} = F_{\text{applied}} - f_k \),
\( F_{\text{net}} = 750 \, N - 392 \, N = 358 \, N \)

Now we can find the acceleration of the crate:
\( F_{\text{net}} = m \cdot a \),
\( 358 \, N = 250 \, kg \cdot a \),
\( a = \frac{358}{250} = 1.43 \, m/s^2 \)

Therefore, the acceleration of the crate is \( 1.43 \, m/s^2 \) when the 750-N force is maintained after the crate begins to move.