This problem has four parts. I will list the question to each part and additional info in "Description" below.

Question

A block of mass m = 150 kg rests against a spring with a spring constant of k = 830 N/m on an inclined plane which makes an angle of θ degrees with the horizontal. Assume the spring has been compressed a distance d from its neutral position.

Part a) Set your coordinates to have the x-axis along the surface of the plane, with up the plane as positive, and the y-axis normal to the plane, with out of the plane as positive. What is an expression for the normal force, F_N, that the plane exerts on the block (in the y-direction) in terms of defined quantities and g?

Part b) Denoting the coefficient of static friction by µ_s, write an expression for the sum of the forces in the x-direction just before the block begins to slide up the inclined plane. Use defined quantities and g in the expression.

Part c) Assuming the plane is frictionless, what will the angle of the plane be, in degrees, if the spring is compressed by gravity a distance 0.1m?

Part d) Assuming θ = 45 degrees and the surface is frictionless, how far will the spring be compressed, d in meters?

Answer 1

An FBD

W_x = –mg sinθ, W_y = –mg cosθ

a) F_N = mg cosθ = (150)(9.8) cosθ = 1470 cosθ N

b) ∑F = F_sp – mg sinθ – μ_sF_N = –kd – mg sinθ – μ_smg cosθ = –kd – mg(sinθ + μ_scosθ)

∑F = 0 just before the block starts to slide, so we can solve for d:

d = –(mg/k)(sinθ + μ_scosθ) = –(1470/830)(sinθ + μ_scosθ) = –1.77(sinθ + μ_scosθ) m

c) F_sp = –kd = mg sinθ. If d = –0.1 m, then

θ = sin^-1(–kd/mg) = sin^-1(83/1470) = 3.24°

d) d = –(mg/k)sinθ = –(1470/830)sin(45°) = –[1470/(830√2)] = –1.25 m

Answer 2

Let's go through each part of the problem step by step:

Part a) To find the expression for the normal force (FN) that the plane exerts on the block in the y-direction, we'll use Newton's second law in the y-direction, which states that the sum of the forces in the y-direction is equal to the mass times acceleration in the y-direction (which is 0 since the block is not moving vertically). The only vertical force is the normal force FN, which acts in the positive y-direction. Therefore:

ΣF_y = FN - mg = 0

Where:

FN = Normal force (upward)

m = Mass of the block = 150 kg

g = Acceleration due to gravity = 9.8 m/s²

Solving for FN:

FN = mg

So, the expression for the normal force FN is:

FN = 150 kg * 9.8 m/s² = 1470 N

Part b) To write an expression for the sum of the forces in the x-direction just before the block begins to slide up the inclined plane, you need to consider the forces acting in that direction. The main forces are the spring force and friction.

Before sliding begins, the maximum static friction force is acting in the opposite direction to motion. The static friction force (Fs) can be calculated as:

Fs = µs * FN

Where:

µs = Coefficient of static friction (given or to be determined)

FN = Normal force (from part a) = 1470 N

The spring force (Fspring) can be calculated using Hooke's law:

Fspring = -k * d

Where:

k = Spring constant = 830 N/m (given)

d = Compression of the spring (given or to be determined)

The sum of forces in the x-direction is zero when the block is on the verge of sliding, so:

ΣF_x = Fs + Fsring = 0

Substituting the expressions for Fs and Fsring:

µs * FN - k * d = 0

Now, rearrange to solve for µs:

µs = (k * d) / FN

Part c) If the plane is frictionless, there is no static friction force acting on the block. In this case, the block will start sliding when the component of the gravitational force parallel to the plane exceeds the maximum static friction force. The component of the gravitational force parallel to the plane is given by:

F_parallel = mg * sin(θ)

Where:

θ = Angle of the plane (to be determined)

m = Mass of the block = 150 kg

g = Acceleration due to gravity = 9.8 m/s²

So, for sliding to occur:

F_parallel > µs * FN (maximum static friction)

Since we're assuming the plane is frictionless, µs = 0, and the equation becomes:

F_parallel > 0

Therefore, the angle θ of the plane can be any angle, and the block will start sliding.

Part d) If θ = 45 degrees and the surface is frictionless, we can find how far the spring will be compressed, d, using energy conservation. The potential energy stored in the compressed spring is converted into gravitational potential energy when the block is lifted up the inclined plane.

The potential energy stored in the spring is given by:

PE_spring = 1/2 * k * d^2

The change in gravitational potential energy as the block is lifted up the inclined plane is given by:

ΔPE_gravity = m * g * h

Where:

h = Height the block is lifted

Since the block moves along the inclined plane, h can be expressed in terms of d and θ:

h = d * sin(θ)

Now, equate the potential energy stored in the spring to the change in gravitational potential energy:

1/2 * k * d^2 = m * g * d * sin(θ)

Plug in the values:

1/2 * 830 N/m * d^2 = 150 kg * 9.8 m/s² * d * sin(45°)

Now, solve for d:

415d^2 = 1470d

Divide both sides by d:

415d = 1470

Now, solve for d:

d = 1470 / 415 ≈ 3.54 meters

So, when θ = 45 degrees and the surface is frictionless, the spring will be compressed by approximately 3.54 meters.