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?
.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
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!!
Let's check your work step-by-step to find the friction force between the crate and the floor.
1. The momentum of the bullet can be calculated using the formula p = mv, where m is the mass and v is the velocity.
Given: mass of the bullet (m) = 6.25 g = 0.00625 kg
velocity of the bullet (v) = 365 m/s
Momentum of the bullet = 0.00625 kg * 365 m/s = 2.28125 N•s (which matches your calculation)
2. After the bullet enters the crate, the crate and bullet move together. To calculate the speed of the crate/bullet system immediately after the collision, you can use conservation of momentum.
Initial momentum = final momentum
(Momentum of bullet before collision) + (Momentum of crate before collision) = (Momentum of system after collision)
Given: mass of the bullet (m_bullet) = 0.00625 kg
velocity of the bullet (v_bullet) = 365 m/s
mass of the crate (m_crate) = 4.50 kg
initial velocity of the crate (v_crate) = 0 (since it was initially at rest)
Initial momentum = (0.00625 kg * 365 m/s) + (4.50 kg * 0 m/s)
= 2.28125 N•s
Mass of system = mass of bullet + mass of crate
= 0.00625 kg + 4.50 kg
= 4.50625 kg
Final velocity of the system (v_system) = initial momentum / mass of system
= 2.28125 N•s / 4.50625 kg
= 0.506 m/s (which matches your calculation)
3. To find the change in kinetic energy (ΔKE) after the collision, you can use the equation ΔKE = 0.5 * m * (v_final^2 - v_initial^2).
Given: mass of the crate (m_crate) = 4.50 kg
initial velocity of the crate (v_initial) = 0 m/s
final velocity of the crate (v_final) = 0.506 m/s
ΔKE = 0.5 * 4.50 kg * ((0.506 m/s)^2 - (0 m/s)^2)
= 0.5 * 4.50 kg * (0.256036 m^2/s^2)
= 0.585225 J (which matches your calculation)
4. The friction force between the crate and the floor can be found using the equation ΔKE = F * d, where F is the friction force and d is the distance.
Given: ΔKE = 0.585225 J
d = 0.15 m (distance the crate slides)
Friction force (F) = ΔKE / d
= 0.585225 J / 0.15 m
= 3.9015 N
Therefore, the friction force between the crate and the floor is approximately 3.9015 N.
Let's go step by step to check your work and find the friction force:
1. You used the formula p = mv to find the momentum. The mass of the bullet is given as 6.25 grams, which is equal to 0.00625 kg. The velocity of the bullet is given as 365 m/s. So the momentum of the bullet is:
p = (0.00625 kg) * (365 m/s) = 2.28125 N·s
Your calculation for momentum seems correct.
2. To find the speed of the crate/bullet system immediately after the collision, you need to consider conservation of momentum. Since there are no external forces acting on the system, the momentum before the collision should be equal to the momentum after the collision.
Initially, only the bullet has momentum, so the momentum is 2.28125 N·s. After the collision, both the bullet and the crate move together. Let's assume the speed of the system is v. The total mass of the system is the mass of the bullet (0.00625 kg) plus the mass of the crate (4.50 kg). So the momentum after the collision is:
p = (0.00625 kg + 4.50 kg) * v
Using the conservation of momentum, we equate the two expressions for momentum:
2.28125 N·s = (0.00625 kg + 4.50 kg) * v
Solving for v:
v = 2.28125 N·s / (0.00625 kg + 4.50 kg) = 0.506 m/s
Your calculation for the speed of the system after the collision is correct.
3. Now let's find the change in kinetic energy. The initial kinetic energy of the system is given by the kinetic energy of the bullet, which can be calculated using the formula KE = (1/2)mv^2. Substitute the mass of the bullet (0.00625 kg) and the speed of the system after the collision (0.506 m/s):
KE_initial = (1/2)(0.00625 kg)(0.506 m/s)^2 = 0.00079795 J
The final kinetic energy of the system is zero since the crate comes to rest. So the change in kinetic energy is:
ΔKE = KE_final - KE_initial = 0 - 0.00079795 J = -0.00079795 J
Your calculation for the change in kinetic energy seems correct, but note that it is negative because the system loses kinetic energy.
4. To find the friction force between the crate and the floor, we can use the work-energy principle. The work done by the friction force is equal to the change in kinetic energy. Therefore, we have:
Work_friction = ΔKE = -0.00079795 J
Recall that work is defined as the product of force and displacement:
Work_friction = Friction * displacement
In this case, the displacement is given as 0.15 m. So we can solve for the friction force:
Friction * 0.15 m = -0.00079795 J
Friction = (-0.00079795 J) / (0.15 m)
Friction = -0.00531967 N
Note that the negative sign indicates that the friction force acts in the opposite direction of the displacement.
Your calculation for the friction force seems correct. However, note that the friction force is negative, indicating that it is acting in the opposite direction of motion.
Therefore, the friction force between the crate and the floor is approximately -0.00532 N.