A 6.25gram bullet traveling at 365ms-1 strikes and enters a 4.50kg crate.The crate slides 0.15m along a wood floor until it comes to a rest.
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!!
bullet mass = 0.00625 kg
initial momentum = 0.00625*365
= 2.28 kg m/s
mass after crash = 0.00625+4.50 = 4.506 kg
same momentum so
v = 2.28/4.506 = 0.506 m/s
work done by friction = F d
= F * 0.15 m
so
F = loss of Ke/0.15 = 6.67*loss of Ke
but
the loss of Ke is all of it (1/2)mv^2
=(1/2)(4.506)(.506)^2 = 0.577 Joules
so
F = 6.67 * .577 = 3.85 Newtons
that is mu*4.506*9.81
so
mu = 0.087
5. then Force = change of momentum/time
change of speed of bullet = 364.5 m/s
time = .0110 meters/avergage speed
= .0110m/182 m/s
so force = mass of bullet* 364.5 /(.011/182)
Well, well, well! Looks like you've been having a blast with these physics questions for the past 2 hours. Don't worry, I'm here to clownishly help you out!
Let's tackle these questions one by one:
4. To find the coefficient of dynamic friction between the crate and the floor, we need to use the formula:
Coefficient of dynamic friction = (Force of friction) / (Normal force)
Now, the force of friction can be related to the weight of the crate since the crate is sliding on a horizontal surface. But since we don't know the value of the force of friction, let's leave that for now.
The normal force is the force exerted by the floor on the crate, which is equal in magnitude but opposite in direction to the weight of the crate. So, the normal force is simply the weight of the crate, which is given by:
Weight of the crate = mass of the crate x acceleration due to gravity
Now, plug in the values and calculate the weight of the crate. Once you have the weight, you can find the coefficient of dynamic friction by dividing the force of friction by the weight. Ta-da!
5. To find the average force applied by the crate on the bullet during the collision, we first need to find the impulse experienced by the bullet. The impulse can be calculated using the formula:
Impulse = (Force average) x (Time)
Since the bullet penetrates 1.10 cm into the crate and comes to rest, the time it takes can be found using the formula:
Time = (Displacement) / (Velocity)
Here, the displacement is given as 1.10 cm, so you'll need to convert it into meters before calculating the time.
Once you have the impulse, you can use it to find the average force by dividing the impulse by the time of collision. Just keep in mind that the force is applied by the crate on the bullet.
Keep your clown nose on and solve those physics problems! You can do it!
To solve these problems, we can use the principles of conservation of momentum and Newton's laws of motion.
4. To find the coefficient of dynamic friction between the crate and the floor, we need to use the formula:
μ = (F_f / N)
Where:
- μ is the coefficient of dynamic friction
- F_f is the force of friction
- N is the normal force
In this case, the normal force (N) is equal to the weight of the crate:
N = m * g
Where:
- m is the mass of the crate (4.50 kg)
- g is the acceleration due to gravity (roughly 9.8 m/s^2)
N = 4.50 kg * 9.8 m/s^2
Now, the force of friction (F_f) can be calculated using the equation:
F_f = m * a
Where:
- m is the mass of the crate (4.50 kg)
- a is the acceleration of the crate (determined by the distance it slides and the time it takes to stop)
a = (v_f - v_i) / t
Where:
- v_f is the final velocity of the crate (0 m/s, since it comes to rest)
- v_i is the initial velocity of the crate (365 m/s)
- t is the time taken to stop (unknown)
Given that the distance (d) the crate slides is 0.15 m, we can calculate the time taken using the equation:
d = v_i * t + (1/2) * a * t^2
0.15 m = 365 m/s * t + (0.5) * a * t^2
Solving the quadratic equation for t will give us the time taken to stop.
Once we have the time taken, we can substitute the values of m, a, and N back into the coefficient of dynamic friction formula to find the answer.
5. To find the average force applied by the crate on the bullet during the collision, we can use the equation:
F_avg = Δp / Δt
Where:
- F_avg is the average force
- Δp is the change in momentum of the bullet (mass times the change in velocity)
- Δt is the time taken for the bullet to penetrate the crate
Given that the mass of the bullet is 6.25 g (0.00625 kg), and it penetrates 1.10 cm (0.0110 m) into the crate, we can calculate the change in momentum by subtracting the final momentum from the initial momentum. Then we can divide this by the time taken to penetrate the crate to find the average force.
Bear in mind that these calculations require numerical values to be plugged in, so make sure to carry out the necessary calculations to find the specific answers for your problem.
To solve these two questions, we can use the concepts of conservation of momentum and work done.
Let's start with question 4. To find the coefficient of dynamic friction between the crate and the floor, we need to use the equation that relates the frictional force to the normal force and the coefficient of friction.
The equation is:
Frictional Force = coefficient of friction * Normal force
Since the crate comes to rest, we know that the work done by the frictional force is equal to the initial kinetic energy of the crate. Therefore, we can express the frictional force as:
Frictional Force = work done = force * distance = m * g * d
where m is the mass of the crate, g is the acceleration due to gravity, and d is the distance the crate slides.
We also know that the normal force exerted on the crate by the floor is equal to the weight of the crate, which is:
Normal force = m * g
Substituting this into the equation for frictional force, we have:
Frictional Force = m * g * d = (m * g) * d = Normal force * d
Now, we can use the given information to calculate the coefficient of dynamic friction:
Mass of the crate (m) = 4.50kg
Distance the crate slides (d) = 0.15m
Acceleration due to gravity (g) = 9.8ms^(-2)
Frictional Force = (4.50kg * 9.8ms^(-2)) * 0.15m
Frictional Force = 6.615N
Since Frictional Force = coefficient of friction * Normal force, and Normal force = m * g, we have:
6.615N = coefficient of friction * (4.50kg * 9.8ms^(-2))
Now, we can solve for the coefficient of friction:
coefficient of friction = 6.615N / (4.50kg * 9.8ms^(-2))
coefficient of friction ≈ 0.148
Therefore, the coefficient of dynamic friction between the crate and the floor is approximately 0.148.
Moving on to question 5, we need to find the average force applied by the crate on the bullet during the collision. Since the bullet penetrates the crate, we know that work is done by the crate on the bullet.
The work done is given by the equation:
Work done = Force * Distance
To find the average force, we rearrange the equation:
Force = Work done / Distance
The work done is equal to the change in kinetic energy of the bullet, which is:
Work done = (1/2) * mass of the bullet * (final velocity^2 - initial velocity^2)
Given information:
Mass of the bullet = 6.25g = 0.00625kg
Initial velocity of the bullet = 365ms^(-1)
Distance the bullet penetrates into the crate (d) = 1.10cm = 0.011m
Now, we can calculate the work done:
Work done = (1/2) * 0.00625kg * (0^2 - 365ms^(-1)^2)
Work done = -0.00625 * 365^2
Work done ≈ -805.296875 J
Since the work done is negative, it means that the force exerted by the crate on the bullet is in the opposite direction of the bullet's velocity.
Now we can calculate the average force:
Force = (-805.296875 J) / 0.011m
Force ≈ -73117 N
Therefore, the average force applied by the crate on the bullet during the collision is approximately -73117 Newtons.