Question

A 3.25-gram bullet traveling at 345 ms-1 strikes and enters a 2.50-kg crate. The crate slides 0.75 m along a wood floor until it comes to rest.

Required:
a. What is the coefficient of dynamic friction between crate and the floor?
b. What is the average force applied by the crate on the bullet during collision if the bullet penetrates the 1.10cm into the crate?

Answer 1

The coefficient of dynamic friction between the crate and the wood floor can be calculated using the work-energy principle by equating the work done by friction to the negative change in kinetic energy. The average force applied by the crate on the bullet during the collision can be estimated using the impulse-momentum theorem and penetration depth.

Step-by-step explanation:

To determine the coefficient of dynamic friction between the crate and the floor, we can use the work-energy principle. The work done by friction is equal to the change in kinetic energy of the system. Since the crate comes to rest, all its kinetic energy is dissipated by the work of friction:

Work done by friction (W) = -Change in kinetic energy (KE)

W = Frictional force (f) × distance (d)

KE = (1/2)mv2

The frictional force is given by f = μ × normal force (N)

In this case, the normal force (N) is the weight of the crate, which is mass (m) × gravity (g).

By equating the work done by friction to the negative change in initial kinetic energy, and solving for the coefficient of kinetic friction (μ), we get:

μ = (-(1/2)mv2)/(mgd)

With m = 2.50 kg, v = 345 ms-1, g = 9.81 ms-2, and d = 0.75 m, we can calculate the coefficient of dynamic friction.

To find the average force applied by the crate on the bullet during the collision, we use the impulse-momentum theorem, which states that the impulse on the bullet equals the change in momentum of the bullet:

Impulse = Force (F) × time (t)

Impulse = Change in momentum = mv - 0, since the bullet comes to rest inside the crate. Instead of time, we can use the penetration depth and bullet speed relation (assuming constant deceleration of the bullet) to find the time by differentiating distance with respect to velocity.

Thus, the average force (F) on the bullet by the crate can be estimated knowing the bullet's mass (m), initial velocity (v), and penetration depth (s).

Answer 2

a) μ = 0.0136, b) F = 22.8 N

Step-by-step explanation:

This exercise must be solved in parts. Let's start by using conservation of moment.

a) We define a system formed by the downward and the box, therefore the forces during the collision are internal and the momentum is conserved

initial instant. Before the crash

p₀ = m v₀

final instant. After inelastic shock

p_f = (m + M) v

the moment is preserved

p₀ = p_f

m v₀ = (m + M) v

v =
`(m)/(m + M) \ v_o`

We look for the speed of the block with the bullet inside

v =
`(0.00325)/(0.00325 + 2.50 ) \ 345`

v = 0.448 m / s

Now we use the relationship between work and kinetic energy for the block with the bullet

in this journey the force that acts is the friction

W = ΔK

W = ½ (m + M)
`v_f^2` - ½ (m + M) v₀²

the final speed of the block is zero

the work between the friction force and the displacement is negative, because the friction always opposes the displacement

W = - fr x

we substitute

- fr x = 0 - ½ (m + M) vo²

fr = ½ (m + M) v₀² / x

the friction force is

fr = μ N

μ = fr / N

equilibrium condition

N - W = 0

N = W

N = (m + M) g

we substitute

μ = ½ v₀² / x g

we calculate

μ = ½ 0.448 ^ 2 / 0.75 9.8

μ = 0.0136

b) Let's use the relationship between work and the variation of the kinetic energy of the block

W = ΔK

initial block velocity is zero vo = 0

F x₁ = ½ M v² - 0

F =
`(1)/(2) M (x)/(y) (v^2)/(x1)`

F = ½ 2.50 0.448² / 0.0110

F = 22.8 N