Question:

A 1.5 kg box is placed on a horizontal surface. The coefficient of friction between the box and the surface is 0.3. The box is attached to a spring with spring constant 500 N/m, as shown in the diagram.

(a) Starting from rest, a constant force of 20 N is exerted on the box, pulling it horizontally. Determine the acceleration of the box.

(b) Determine the maximum compression of the spring when the box comes to rest.

(c) If instead of a constant force, a constant acceleration of 5 m/s^2 is applied to the box, what is the tension in the spring when the box comes to rest?

Assume no energy loss due to friction, and neglect the mass of the spring itself.

Answer:

(a) To find the acceleration of the box, we need to use Newton's second law, which states that the net force acting on an object is equal to the mass of the object times its acceleration.

The net force acting on the box is the force applied minus the force of friction. The force of friction can be calculated using the equation:

Frictional force (Ff) = coefficient of friction (µ) × normal force (Fn)

The normal force is equal to the weight of the box, which can be calculated as:

Weight (W) = mass (m) × gravity (g)

Given: Force applied (F_applied) = 20 N Coefficient of friction (µ) = 0.3 Mass of the box (m) = 1.5 kg Gravity (g) = 9.8 m/s^2

First, calculate the weight of the box: Weight (W) = 1.5 kg × 9.8 m/s^2 = 14.7 N

Next, calculate the frictional force: Frictional force (Ff) = 0.3 × 14.7 N = 4.41 N

Net force (F_net) = F_applied - Ff F_net = 20 N - 4.41 N = 15.59 N

Now, use Newton's second law to find the acceleration: F_net = m × a 15.59 N = 1.5 kg × a

Solving for acceleration (a): a = 15.59 N / 1.5 kg = 10.39 m/s^2

Therefore, the acceleration of the box is 10.39 m/s^2.

(b) To determine the maximum compression of the spring when the box comes to rest, we can use the spring potential energy formula:

Potential energy (PE) = (1/2) × k × x^2

Where k is the spring constant and x is the compression of the spring.

When the box comes to rest, its kinetic energy will be zero. Therefore, all of the initial energy from the applied force is stored as potential energy in the spring.

Potential energy (PE) = Force applied (F_applied) × distance (d)

Given that the force applied is 20 N, we need to find the distance d traveled by the box. We can use the equation:

Distance (d) = (v_f^2 - v_i^2) / (2 × a)

Where v_f is the final velocity (0 m/s since the box comes to rest) and v_i is the initial velocity (0 m/s since the box starts from rest).

Plugging in the values: d = (0^2 - 0^2) / (2 × 10.39 m/s^2) = 0 m

Therefore, the maximum compression of the spring is 0 m when the box comes to rest.

(c) If a constant acceleration of 5 m/s^2 is applied to the box, then the net force can be calculated using Newton's second law:

F_net = m × a F_net = 1.5 kg × 5 m/s^2 = 7.5 N

However, the net force is not equal to the tension in the spring. The tension in the spring can be found by considering the forces acting on the box. When the box comes to rest, the force of friction will be equal to the net force.

Frictional force (Ff) = F_net = 7.5 N

The frictional force can be calculated using the equation:

Frictional force (Ff) = coefficient of friction (µ) × normal force (Fn)

The normal force is equal to the weight of the box, which we previously calculated as 14.7 N.

14.7 N = µ × Fn 14.7 N = 0.3 × Fn

Solving for the normal force (Fn): Fn = 14.7 N / 0.3 = 49 N

Therefore, the tension in the spring when the box comes to rest is equal to the normal force, given as 49 N.