Question

A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m the floor is frictionless, and for the next 10.5 m the coefficient of friction is 0.17.

What is the final speed of the crate after being pulled these 20.5 meters?

Answer 1

The final speed of the crate is 12.07 m/s.

Step-by-step explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

`F = ma`

`a = (F)/(m) = (225 N)/(51.0 kg) = 4.41 m/s^(2)`

Now, we can calculate the final speed of the crate at the end of 10.0 m:

`v_(f)^(2) = v_(0)^(2) + 2ad_(1)`

`v_(f) = \sqrt{0 + 2*4.41 m/s^(2)*10.0 m} = 9.39 m/s`

For the next 10.5 meters we have frictional force:

`F - F_(\mu) = ma`

`F - \mu mg = ma`

So, the acceleration is:

`a = (F - \mu mg)/(m) = (225 N - 0.17*51.0 kg*9.81 m/s^(2))/(51.0 kg) = 2.74 m/s^(2)`

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

`v_(f)^(2) = v_(0)^(2) + 2ad_(2)`

`v_(f) = \sqrt{(9.39 m/s)^(2) + 2*2.74 m/s^(2)*10.5 m} = 12.07 m/s`

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.

I hope it helps you!