Question

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?

Answer 1

Answer:

12 m/s

Explanation:

The acceleration in the first 15 m is

$a_1 = F / m = 350 / 96 = 3.646 m/s^2$

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

$v_1^2 - v_0^2 = 2a\Delta s$

where v1 velocity of the crate at the end of the 1st 15 m, $v_0$ 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 $\Delta s = 15$ is the distance traveled.

$v_1^2 = 2*3.646*15 = 109.375$

$v_1 = \sqrt{109.375} = 10.46 m/s$

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

$F_f = \mu N = \mu mg = 0.25 * 96 * 9.81 = 235.44 N$

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

$v^2 - v^2 = 2a_2\Delta s$

$v^2 - 109.375 = 2* 1.19*15 = 35.7$

$v^2 = 145.045$

$v = \sqrt{145.045} = 12 m/s$