Question

A box of mass m is initially held at rest (at point A) on an incline forming an angle θ with the horizontal. When released, it slides a distance d down the incline which has a rough surface with a coefficient of kinetic friction µk. After it reaches the bottom of the incline (point B), it continues to slide on a horizontal icy, frictionless surface. At point C, the box sticks to the end of a spring which has a negligible mass and spring constant k. The box compresses the spring as it slows down until the spring reaches its maximum compression. Express your answers to this problem in terms of the acceleration due to gravity g and the quantities defined above: m, θ, d, µk and k.

Answer 1

`v_(f)` = √[2g (sin θ - μ cos θ) d ] and x = √ [2g d m/k (sin θ - μ cos θ)]

Step-by-step explanation:

We must solve this problem in parts, let's start by working with points A and B, we can use the principles of energy, in this case there is friction so let's use Newton's second law to find the acceleration of the body and with it the speed in the lowest point of the plane

The reference system has the x axis parallel to the plane and the axis and perpendicular, the weight is the only force that we must decompose

sin θ = Wₓ / W

cos θ =
`W_(y)` / W

Wₓ = W sin θ

`W_(y)` = W cos θ

We write the equations on each axis

Y Axis

N-
`W_(y)` = 0

N =
`W_(y)`

N = mg cos θ

X axis

Wₓ -fr = m a

Friction force is

fr = μ N

mg sin θ - μ mg cos θ = ma

a = g (sin θ - μ cos θ)

This is the acceleration along the plane. Let's use the kinematic equations to find the velocity in the lower part of the plane, as the body is initially at rest its zero velocity (vo = 0) and the distance traveled is the length of the plane (d)

`v_(f)`² = v₀² + 2 a d

`v_(f)`² = 0 + 2 a d

`v_(f)` = √[2g (sin θ - μ cos θ) d ]

We already have the speed at point B

Now at point C it hits a spring and compresses it, we can use energy conservation to find the compression of the spring, let's write the mechanical energy at points C and D

Initial point C

Em₀ = K = ½ m v²

Final point D

Em₁ = Ke = ½ k x²

Em₀ = Em₁

½ m v² = ½ k x²

x² = m / k v²

x² = m / k 2 g (sin θ - μ cos θ) d

x = √ [2g d m/k (sin θ - μ cos θ)]