Question

A 6.25-gram bullet travels at 365 m/s. It hits and enters a 4.50-kg crate. The crates slides 0.15m along a floor until it comes to rest.
Find the friction force between the crate and the floor

I used p=mv to get momentum and got 2.28125 N•s
I calculated speed of the crate/bullet system immedatiately after the collision and got 0.506 m/s
I also had to find the change in kinetic energy after the collision, and got 416.328125 J - 0.585225 = 415.7429 J

Can someone please check my work and tell me how to find the Friction force?

Answer 1

.00625 *365 = 4.50625 v
v= .506 m/s yes

average speed during stop = .506/2 = .253 m/s

time to stop = .15 / .253 = .593 second
acceleration = change in speed/time
= -.506 /.593 = -.854 m/s^2

F = m a = 4.50625 *-.854
= -3.85 Newtons

Ke initial = .5 * .00625 * 365^2
= 416 J
Ke final = .5 * 4.50625 * .506^2
=.577 J

Answer 2

could someone continue solving these following questions based on the same question the above person posted?
4. What is the coefficient of dynamic friction between crate and the floor?
5. What is the average force applied by the crate on the bullet during collision if the bullet penetrates the 1.10cm into the crate?
Please help asap I've been stuck with these two question for honestly about 2 hours!!