Physics 11- Friction

Question

A crate of mass 65.7kg is being pushed across a concret floor with a constant force of 499N. How long will it take to push the crate a distance of 28.8m? The coefficient of kinetic friction is 0.750.

Answer 1

Always start out with a free body diagram.

If we're pushing the crate to the right, we have:

1) 499N applied force to the right

2) Frictional force to the left

3) Gravitational force down

4) Normal force up

The gravitational force and the normal force cancel out since the crate isn't accelerating vertically

Therefore the normal force equals the gravitational force

So F_norm = F_g = (65.7kg)(9.8m/s²) = 643.86N

F_fric = F_norm*µ = 643.86N * 0.75 = 482.895N

Now we use the equation ∑F_net = ma since figuring out how fast the crate is accelerating will tell us how long it took it to slide 28.8m. The greater the acceleration, the less time it will take.

Here we only care about the forces and acceleration in the horizontal direction

The forces in the horizontal direction are the applied force and the frictional force

The applied force is to the right so that's a positive force, and the frictional force is to the left, so that's a negative force. The block is going to be accelerating to the right, so acceleration should come out positive.

F_applied - F_fric = ma

(499N) - (482.895N) = (65.7kg)*a

16.105N = (65.7kg)* a

a = 0.245m/s²

Now, you have to find time. We're given the initial speed, the acceleration, and the distance, and we need to find the time taken. So the kinematic equation we would use is x = 1/2 at² + v₀t.

Now, plug in and solve for t.