Question

1. A 25 kg box is lifted upward at a constant rate of 2 m/s. If the box is lifted a height of 4 m, how much work is done on the box?

2. A 120 kg box is attached to a string and pulled along a rough, flat surface with a coefficient of friction 0.22. If the string pulls with a force of 500 N and is attached at an angle of 30 degrees above the horizontal, and the box moves strictly horizontally a distance of 5 m.

How much work is done on the box by pulling on the string?

3.

A 120 kg box is attached to a string and pulled along a rough, flat surface with a coefficient of friction 0.22. If the string pulls with a force of 500 N and is attached at an angle of 30 degrees above the horizontal, and the box moves strictly horizontally a distance of 5 m.

How much work is done on the box by gravity?

4.

A 120 kg box is attached to a string and pulled along a rough, flat surface with a coefficient of friction 0.22. If the string pulls with a force of 500 N and is attached at an angle of 30 degrees above the horizontal, and the box moves strictly horizontally a distance of 5 m.

How much work is done on the box by friction?

4.

A 120 kg box is attached to a string and pulled along a rough, flat surface with a coefficient of friction 0.22. If the string pulls with a force of 500 N and is attached at an angle of 30 degrees above the horizontal, and the box moves strictly horizontally a distance of 5 m.

How much work is done on the box by friction?

4.

A 120 kg box is attached to a string and pulled along a rough, flat surface with a coefficient of friction 0.22. If the string pulls with a force of 500 N and is attached at an angle of 30 degrees above the horizontal, and the box moves strictly horizontally a distance of 5 m.

How much work is done on the box by friction?

A 120 kg box is attached to a string and pulled along a rough, flat surface with a coefficient of friction 0.22. If the string pulls with a force of 500 N and is attached at an angle of 30 degrees above the horizontal, and the box moves strictly horizontally a distance of 5 m.

How much work is done on the box by friction?

A 120 kg box is attached to a string and pulled along a rough, flat surface with a coefficient of friction 0.22. If the string pulls with a force of 500 N and is attached at an angle of 30 degrees above the horizontal, and the box moves strictly horizontally a distance of 5 m.

How much work is done on the box by friction?

A 120 kg box is attached to a string and pulled along a rough, flat surface with a coefficient of friction 0.22. If the string pulls with a force of 500 N and is attached at an angle of 30 degrees above the horizontal, and the box moves strictly horizontally a distance of 5 m.

How much work is done on the box by friction?

4. A 120 kg box is attached to a string and pulled along a rough, flat surface with a coefficient of friction 0.22. If the string pulls with a force of 500 N and is attached at an angle of 30 degrees above the horizontal, and the box moves strictly horizontally a distance of 5 m.

How much work is done on the box by friction?

5. In  the previous problem concerning the work done by friction. The work done by friction would be negative as it is removing energy from the system and working to slow the motion. True or False?

6. In the previous problems, the Normal force opposes gravity and prevents the object from falling through the surface it is sliding across. The work done by the normal force in this case is negative as it is stopping the motion in the vertical direction. True or False?

Answer 1

Multiple questions posted, so answering only the first question:

1. Since the object is moving with constant velocity, there is no acceleration. From newton's second law, we know that : force=mass*acceleration.Since, acceleration=0, force=0 newtons.So,there is no net force acting on the object.

Now forces on the object include:

a)gravitational force =mg downwards, where m is mass and g is gravitational acceleration.

b)force applied by the external agent=F newtons upwards.

Since there is no net force acting on the object, F=mg.

Now,work done is given by: W=F*s*cos, where W is work done, F is magnitude of force vector, s is magnitude of displacement vector, and is the angle between the force and displacement vectors.

Here,F=mg. In the given problem , m=25 kg. So, F=25*9.8=245 N., upwards

Displacement=4 m upwards

=0 degrees, as both force and displacement vector are in the same direction,i.e., upward.

So,work done=245*4*cos0=980 J.