How Do Friction and Human Force Affect Work Done on a Box?

[gcharles_42]

Homework Statement

A box of mass 45 kg is initially at rest on a flat floor. The coefficient of kinetic friction between the box and the floor is 0.60. A woman pushes horizontally against the box with a force of 539 N until the box attains a speed of 2 m/s.

What is the CM-work done by the friction force on the box?

What is the work done by the woman on the box?

Homework Equations

W= F (Δ x)

The Attempt at a Solution

I tried solving for delta x by using force and mass to calculate for acceleration and then acceleration, Vo and Vf to find d (which I thought was Δ x) with the equation vf^2 = vo^2 + 2(a) d but I got the wrong answer

 

[STEMucator]

Homework Statement

A box of mass 45 kg is initially at rest on a flat floor. The coefficient of kinetic friction between the box and the floor is 0.60. A woman pushes horizontally against the box with a force of 539 N until the box attains a speed of 2 m/s.

What is the CM-work done by the friction force on the box?

What is the work done by the woman on the box?

Homework Equations

W= F (Δ x)

The Attempt at a Solution

I tried solving for delta x by using force and mass to calculate for acceleration and then acceleration, Vo and Vf to find d (which I thought was Δ x) with the equation vf^2 = vo^2 + 2(a) d but I got the wrong answer

You should get in the habit of drawing a free body diagram. They can basically solve the entire problem for you most of the time.

If you do that, you will notice there are multiple forces at work here!

 

[Legaldose]

You should get in the habit of drawing a free body diagram. They can basically solve the entire problem for you most of the time.

If you do that, you will notice there are multiple forces at work here!

This. Free body diagrams are the best thing that ever happened to intro mechanics.

 

[lep11]

Always show your work so we can tell you where it went wrong.

 

[Nickie]

.

I would approach this problem by first defining the concepts of work and friction. Work is defined as the transfer of energy from one object to another, and it is calculated by multiplying the force applied by the distance over which it is applied. Friction, on the other hand, is a force that opposes motion and is caused by the interaction between two surfaces in contact.

In this scenario, the work done by the friction force on the box can be calculated by multiplying the force of friction (which is equal to the coefficient of kinetic friction multiplied by the normal force) by the distance over which it acts. This distance can be calculated by using the equation vf^2 = vo^2 + 2(a) d, where vf is the final velocity (2 m/s), vo is the initial velocity (0 m/s), and a is the acceleration (which can be calculated using the force of friction and the mass of the box). This will give the distance traveled by the box due to the friction force. The work done by the woman on the box can be calculated by multiplying the force she applied (539 N) by the distance over which it was applied (which can be calculated using the same equation as before).

Therefore, the CM-work done by the friction force on the box is equal to the force of friction (μk * mg) multiplied by the distance traveled by the box due to the friction force. The work done by the woman on the box is equal to the force applied (539 N) multiplied by the distance traveled by the box due to the woman's force. It is important to note that in this scenario, the work done by the friction force is negative (since it acts in the opposite direction of motion) while the work done by the woman is positive (since she is causing the box to move in the direction of her force).