A 96-kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 350 N. For the first 15 m the floor is frictionless, and for the next 15 m the coefficient of friction is 0.25. What is the final speed of the crate?

12 m/s

Explanation:

The acceleration in the first 15 m is

The velocity at the end of the first 15 m can be calculated using the following equation of motion:

where v1 velocity of the crate at the end of the 1st 15 m, 0m/s is the initial velocity when starts from rest, a = 3.646 m/s2 is the acceleration of the crate in the 1st 15m, and is the distance traveled.

As for the 2nd 15m, let's find the friction force first. Let's g = 9.81 m/s2

The net force would be 350 - 235.44 = 114.56 N

Then the net acceleration be:

114.56 / 96 = 1.19

Using the similar equation of motion above, we can calculate the final velocity

momentum of soft ball three times less than baskey ball