Question

Chris is in the process of moving to a new house, and he needs to carry out a lot of boxes from the second floor to his pickup truck. The mass of each box is 53 kg. Instead of carrying boxes out one by one, he has set up a smooth, frictionless slope from the second floor to the first floor so that he can slide down boxes one by one. When a box slides down to the first floor, it continues sliding by a distance of 7.8 m toward the entrance of the house, where the pickup truck is parked. There is a small, frictionless ramp connecting to the bed of the pickup truck so the box can be loaded to the truck effortlessly.

See attached image

The first floor is carpeted, and there is a frictional force of magnitude 140 N on the box as it slides on the carpet. The height difference between the first and second floor is 3.2 m, and the height difference between the first floor and the bed of the pickup truck is 0.90 m.

A box is initially at rest on the second floor, and Chris pushes the box toward the slope so that the speed of the box is 2.1 m/s right before it starts sliding down the slope. The second floor is smooth, and the frictional force between the second floor and the box is negligible.

Use g = 10 m/s2 for the acceleration due to gravity.



(1)
What is the work done by Chris on the box when the speed of the box reaches 2.1 m/s?

(2)
What is the speed of the box when it reaches the bottom of the slope (Point B in the diagram)?

(3)
To what speed does the box slow down when it reaches to the bottom of the ramp to the pickup truck?

(4)
What is the speed of the box when it reaches the bed of the pickup truck?

(5)
If instead Chris just pushes the box off the slope from rest (i.e., initial speed is 0 m/s), does the box make it to the bed of the truck? Assume that the magnitude of the frictional force is still 140 N. Show your calculation to support your answer.

Answer 1

To solve the given problems, we'll use the principles of work-energy and conservation of energy. Let's address each question one by one:

(1) What is the work done by Chris on the box when the speed of the box reaches 2.1 m/s?

The work done by Chris on the box is equal to the change in the box's kinetic energy. Since the box starts from rest, the initial kinetic energy is zero. The final kinetic energy can be calculated using the formula:

Kinetic energy = (1/2) * mass * velocity^2

Plugging in the values:

Mass of the box (m) = 53 kg

Final velocity (v) = 2.1 m/s

Kinetic energy = (1/2) * 53 kg * (2.1 m/s)^2

Calculate the value of the kinetic energy, which represents the work done by Chris on the box.

(2) What is the speed of the box when it reaches the bottom of the slope (Point B in the diagram)?

To determine the speed at the bottom of the slope, we'll use the principle of conservation of energy. The total mechanical energy of the box is conserved as it moves from the top to the bottom of the slope.

The initial potential energy at the top of the slope is converted into kinetic energy at the bottom of the slope, neglecting any energy losses due to friction.

Potential energy at the top = m * g * h1

Where:

Mass of the box (m) = 53 kg

Acceleration due to gravity (g) = 10 m/s^2

Height difference between floors (h1) = 3.2 m

Calculate the initial potential energy.

The final kinetic energy at the bottom is given by:

Kinetic energy at the bottom = (1/2) * m * v^2

Where:

Mass of the box (m) = 53 kg

Velocity at the bottom (v) = ?

Equating the initial potential energy to the final kinetic energy, solve for v to find the speed of the box at the bottom of the slope.

(3) To what speed does the box slow down when it reaches the bottom of the ramp to the pickup truck?

Since the ramp connecting the first floor to the bed of the pickup truck is frictionless, there is no external force doing work on the box. Thus, the mechanical energy of the box is conserved as it moves from the bottom of the slope to the bottom of the ramp.

Using the same principle of conservation of energy, equate the final kinetic energy at the bottom of the slope to the initial potential energy at the bottom of the ramp.

Potential energy at the bottom of the ramp = m * g * h2

Where:

Mass of the box (m) = 53 kg

Acceleration due to gravity (g) = 10 m/s^2

Height difference between the first floor and the truck bed (h2) = 0.90 m

Calculate the potential energy at the bottom of the ramp.

Equating the potential energy at the bottom of the ramp to the final kinetic energy, solve for the speed of the box at the bottom of the ramp.

(4) What is the speed of the box when it reaches the bed of the pickup truck?

Since the ramp connecting the first floor to the truck bed is frictionless, there is no external force doing work on the box. The mechanical energy of the box is conserved as it moves from the bottom of the ramp to the truck bed.

Using the same principle of conservation of energy, equate the final potential energy at the bottom of the ramp to the final kinetic energy at the truck bed.

Potential energy at the truck bed = m * g * h