# How to calculate work done by forces on a block moving up a wall?

• AznBoi
In summary, the problem is about a 5kg block being pushed 3m up a vertical wall with constant speed by a force of magnitude F at an angle of 30 degrees with the horizontal. The work done by F is 208 N. The work done by the force of gravity is -58.8 J. The normal force between the block and wall is 0 N. The gravitational potential energy increases by 58.8 J during this motion. The angle between the force and displacement for finding the work done by the force is 60 degrees.
AznBoi
Okay I kind of get this problem but not completey. Please see if I'm going in the right direction and guide me through it. Thanks!

Here is the Problem: A 5kg block is pushed 3m up a vertical wall with constant speed by a constant force of magnitude F applied at an angle of theta=30 degrees with the horizontal. If the coefficient of kinetic friction between block and wall is 0.30, determine the work done by a) F, b) the force of gravity, and c) the normal force between block and wall. d) By how much does the gravitational potential energy increase during this motion?

Here is my work:
First of all I solved for the F force and I got 208 N
So then do I have to plug the force into the equation: W=Fcos(theta)(displacement) ??

Then it would be: 208cos30(3m) right??

For b) you need to find the work force of gravity.
So, W=(9.8m/s^2)cos180(3m)??

for c) normal force between block and wall.
would it be zero?? because the block moves upward and the normal force would be perpendicular to the displacement.

for d) How do you find how much the gravitational potential energy increases by? Do you use Usubi+Ksubi=Usubf+Usuf

You have a problem with the angle. Potential energy increase is the same as always. Your equation is fine.

What do you mean I have a problem with the angle?? Are all of the problems that I attempted right?

I think he is talking about the work done by the force. 30 degrees is not the angle you should use, if I am understanding this problem correctly.

It is only the component of the force in the direction of the motion that is doing any work.

Dorothy

I think 30 degrees is the angle that you need to use for the Fore F because it is being applied diagonally 30 degrees from the horizontal upwards.

Just a suggestion, but a closer look at the definition of the angle in that particular formula would be a good idea. Hint: It's the angle between two vectors.

Dorothy

oh so if the angle is 30 degrees from the horizontal. and the displacement is 90 degrees. would it be 90-30=60 degrees?

That's right.

would this be right??

For b) you need to find the work force of gravity.
So, W=(9.8m/s^2)cos180(3m)??

because the force of gravity acts downards and the displacement is upwards so it would be cos180 right?

Yes, that's right.

## What is work done by forces?

Work done by forces is the measure of the energy transferred to an object by a force, resulting in a displacement of the object in the direction of the force.

## How is work done calculated?

Work done is calculated by multiplying the magnitude of the force applied by the displacement of the object in the direction of the force. The formula is: W = F x d, where W is work done, F is force, and d is displacement.

## What are the units of work done?

The units of work done are joules (J) in the International System of Units (SI). However, work can also be expressed in other units such as foot-pounds (ft-lb) or calories (cal).

## Can work be negative?

Yes, work can be negative. This occurs when the force and displacement are in opposite directions, resulting in a decrease in the object's energy. Negative work is often associated with friction or other forms of resistance.

## What are some real-life examples of work done by forces?

Some examples of work done by forces include lifting a book, pushing a shopping cart, or pulling a suitcase. Work is also done when a car accelerates, when a rocket launches, or when a crane lifts heavy objects.

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