A 1.8 kg breadbox on a frictionless incline of angle θ = 37° is connected, by a cord that runs over a pulley, to a light spring of spring constant k = 120 N/m, as shown in the figure below. The box is released from rest when the spring is unstretched. Assume that the pulley is massless and frictionless.

(a) What is the speed of the box when it moves 8 cm down the incline?
m/s

(b) How far down the incline from its point of release does the box slide before momentarily stopping?
m

(a) What is the speed of the box when it moves 8 cm down the incline?

Well, you know what they say, speed is relative. But let's calculate it anyway!

First, we need to find the net force acting on the box. Since there is no friction, the only forces in play are gravity and the spring force. The component of gravity acting down the incline is mg * sin(37°).

Now let's look at the spring force. The displacement of the spring when the box moves 8 cm down the incline is given by x = 8 cm = 0.08 m. Using Hooke's Law, we can calculate the force exerted by the spring, F_spring = -kx. Remember that the negative sign is there because the spring force is acting in the opposite direction to the displacement.

So the net force will be the sum of the gravitational force and the spring force, and it can be calculated as F_net = mg * sin(37°) + (-k * 0.08).

Next, we'll use Newton's second law, F = ma, to find the acceleration. Since the mass of the box is given as 1.8 kg, the acceleration can be calculated as a = F_net / m.

Finally, we can find the final speed of the box using the equation v^2 = u^2 + 2as, where u is the initial speed of 0 m/s (since the box was at rest) and s is the displacement of 8 cm = 0.08 m.

Now, it's time for some math magic!

F_net = (1.8 kg) * (9.8 m/s^2) * sin(37°) + (-120 N/m * 0.08 m)
= (1.8 kg) * (9.8 m/s^2) * 0.6018 + (-9.6 N)
≈ 10.93 N

a = (10.93 N) / (1.8 kg)
≈ 6.07 m/s^2

v^2 = (0 m/s)^2 + 2 * (6.07 m/s^2) * (0.08 m)
= 0 + 0.9752 m^2/s^2
≈ 0.9752

Taking the square root of both sides, we find that the speed of the box when it moves 8 cm down the incline is approximately:

v ≈ 0.988 m/s

(b) How far down the incline from its point of release does the box slide before momentarily stopping?

Well, if we first find the acceleration of the box as it slides down, we can then use the kinematic equation v^2 = u^2 + 2as to find the distance it travels before stopping.

Since the box is momentarily stopping, we know that the final velocity, v, is 0 m/s. The initial velocity, u, can still be assumed as 0 m/s because the box is released from rest. We'll also let the distance traveled be denoted by s.

Rearranging the equation gives us s = -u^2 / (2a). Plugging in the values we know:

s = -(0 m/s)^2 / (2 * 6.07 m/s^2)
= 0 m

Uh-oh! Looks like it stopped right at the point of release. Maybe the box was craving some nostalgia or wanted to enjoy the view a little longer.

To solve the problem, we will use the concepts of work-energy theorem and Hooke's Law.

Let's break down the problem step by step:

Step 1: Determine the acceleration of the system
The forces acting on the system are gravity, normal force, and tension in the cord. Since the pulley is frictionless, there is no friction force. The net force along the incline can be expressed as:

F_net = m * a
m * g * sin(θ) - T = m * a

where
m = mass of the breadbox = 1.8 kg
g = acceleration due to gravity = 9.8 m/s^2
θ = angle of the incline = 37°
T = tension in the cord
a = acceleration of the system

Using the equation above, we can solve for the acceleration:

1.8 kg * 9.8 m/s^2 * sin(37°) - T = 1.8 kg * a
16.2868 N - T = 1.8 kg * a

Step 2: Determine the tension in the cord
The tension in the cord can be determined by considering the forces acting on the breadbox in the vertical direction:

T - m * g * cos(θ) = 0
T = m * g * cos(θ)

Substituting the values we have:

T = 1.8 kg * 9.8 m/s^2 * cos(37°)
T = 1.8 kg * 7.8603 m/s^2
T = 14.14854 N

Step 3: Determine the acceleration (continued)
Now that we know the tension in the cord, we can substitute this value back into the equation from Step 1 to solve for the acceleration:

16.2868 N - 14.14854 N = 1.8 kg * a
2.13826 N = 1.8 kg * a
a = 1.1879 m/s^2

Step 4: (a) Calculate the speed of the box
Using the kinematic equation for uniform acceleration:

v^2 = u^2 + 2 * a * s

where
v = final velocity
u = initial velocity (which is 0 since the box was released from rest)
a = acceleration
s = displacement

We want to find the final velocity when the box moves 8 cm down the incline. Converting 8 cm to meters:

s = 0.08 m

Plugging in the values:

v^2 = 0 + 2 * 1.1879 m/s^2 * 0.08 m
v^2 = 0.190048 m^2/s^2
v = √(0.190048 m^2/s^2)
v = 0.4364 m/s

Therefore, the speed of the box when it moves 8 cm down the incline is 0.4364 m/s.

Step 5: (b) Determine the distance the box slides before momentarily stopping
When the box momentarily stops, its velocity becomes zero. Using the same kinematic equation as in Step 4, we can solve for the distance:

v^2 = u^2 + 2 * a * s

0 = u^2 + 2 * 1.1879 m/s^2 * s

Since the box is initially released from rest, the initial velocity (u) is 0. Solving for s:

s = -u^2 / (2 * a)
s = -(0^2) / (2 * 1.1879 m/s^2)
s = 0 m

Therefore, the box doesn't slide at all before momentarily stopping.

To find the answers to these questions, we can use the principles of Newton's laws, energy conservation, and Hooke's law.

(a) What is the speed of the box when it moves 8 cm down the incline?

To find the speed of the box, we need to determine the net force acting on it. We can do this by calculating the gravitational force and the force exerted by the spring.

1. Calculate the gravitational force:
The force of gravity acting on the box is given by the equation: F_gravity = m * g * sin(θ).
where m = mass of the box = 1.8 kg, g = acceleration due to gravity = 9.8 m/s^2, and θ = angle of the incline = 37°.

F_gravity = 1.8 kg * 9.8 m/s^2 * sin(37°) = 31.7 N.

2. Calculate the force exerted by the spring:
The force exerted by the spring is given by Hooke's law: F_spring = -k * x.
where k = spring constant = 120 N/m, and x = displacement of the box from its unstretched position.

Since the box is moving down the incline, the displacement x is negative. We need to find the value of x corresponding to a 8 cm displacement.

x = -0.08 m (converted from centimeters to meters).

F_spring = -120 N/m * (-0.08 m) = 9.6 N.

3. Calculate the net force:
The net force is the vector sum of the gravitational force and the force exerted by the spring:

Net force = F_gravity + F_spring = 31.7 N + 9.6 N = 41.3 N.

4. Use Newton's second law to find the acceleration:
According to Newton's second law, the net force is equal to the mass of the object multiplied by its acceleration:

Net force = m * a,
41.3 N = 1.8 kg * a.

a = 41.3 N / 1.8 kg ≈ 22.94 m/s^2.

5. Use kinematic equations to find the speed of the box:
Using the equation for motion in the x-direction:

v^2 = u^2 + 2a * s,
where v = final velocity, u = initial velocity (0 m/s as the box was released from rest), a = acceleration, and s = displacement.

v^2 = 0 + 2 * 22.94 m/s^2 * 0.08 m,
v^2 = 3.6704 m^2/s^2.

v ≈ √(3.6704) ≈ 1.92 m/s.

Therefore, the speed of the box when it moves 8 cm down the incline is approximately 1.92 m/s.

(b) How far down the incline from its point of release does the box slide before momentarily stopping?

To find the distance the box slides before momentarily stopping, we need to use the concept of energy conservation.

1. Calculate the potential energy:
The potential energy of the system is given by the equation: PE = m * g * h,
where m = mass of the box = 1.8 kg, g = acceleration due to gravity = 9.8 m/s^2, and h = height of the incline.

The height of the incline can be calculated using trigonometry:

h = L * sin(θ),
where L = length of the incline, and θ = angle of the incline.

As the box slides down the incline, its potential energy is converted into kinetic energy. At the point of momentarily stopping, all the potential energy is converted to kinetic energy.

2. Calculate the kinetic energy:
The kinetic energy of the box is given by the equation: KE = (1/2) * m * v^2,
where m = mass of the box = 1.8 kg, and v = velocity of the box at the point of momentarily stopping.

At the point of momentarily stopping, the kinetic energy is zero.

3. Set the potential energy equal to zero:
Since all the potential energy is converted to kinetic energy at the point of momentarily stopping, we can set the potential energy equal to zero:

m * g * h = 0.

4. Solve for h:
h = 0 / (m * g) = 0 m.

5. Use the expression for h in terms of L and θ:
h = L * sin(θ),
0 = L * sin(37°).

Since sin(37°) ≠ 0, the length of the incline, L, must be zero for the box to momentarily stop.

Therefore, the box does not slide down the incline from its point of release before momentarily stopping as the length of the incline, L, is zero.