# Who is right - me or the book - work and kinetic energy

• Hemmelig
In summary, the book states that the answer is 99.2N, while the experiment shows that the answer is 84.95N. The book is right.

#### Hemmelig

My educated guess would be that I'm wrong

## Homework Statement

A factory worker pushes a 30kg crate a distance of 4,5m along a floor at constant velocity, the coefficient of kinetic friction between the crate and floor is 0.25

The angle he pushes the crate with is 30 degrees below the horizontal

## Homework Equations

Work = F * s * cos(angle)

Normal force = mg

## The Attempt at a Solution

What i do know is that since the velocity is constant, the total work done on the crate is 0

So Work total =Wt = Work done by worker + Work done on crate by friction = 0

So Work done by worker = - work done on crate by friction

Work done by worker = F*s * cos(30)

work done by friction = F*s * cos(180) = F*s*-1

I find the friction force by using the frictional coefficient * the normal force

So F = 0.25 * 30 * 9.81 = 73,575N

F*s*cos(30)= -73,575N*s*-1

I can remove s from both sides

F*cos(30) = 73,575N

F=73,575/cos30 = 84,95N

Which is the wrong result

The book states that the answer is 99.2N

What am i doing wrong ?

Of course, the book is right.

Now, you don't need the energies here. It's all about forces. Since the crate moves at constant velocity, the net force on it should be zero (1st Newton's law).

You ignored the y component of the force that the crate is being pushed by. It effects the force that crate exerts on the ground (it also effects the friction!).

As Dr Jekyll states above, leave energies out of it. I can't understand why you're given a distance so ignore that also.

I'd like to say that I can't get that answer if the force acts at an angle below the horizontal yet i get the exact answer if I assume it acts at an angle above, so i'll assume that that part of the information is incorrect (?)

Resolve the components of the force acting on the box into y and x.

Fx = F x cos(30)
Fy = F x sin(30)

The y component of the force acts to increase the normal force, R, experienced by the box;

R = Fy + mg

so you can then work out the frictional force using the formula you've mentioned.

Since the box is at a constant speed, this frictional force should be equal to the x component of force acting on the box. You'll be left with an equation with F. Rearrange this to find your answer.

But he's pushing down at the box with an angle of 30 , won't the frictional force be horizontal, as in f x cos (180) then ?

Hemmelig said:
Normal force = mg
Since there is an applied force pushing down, the normal force does not equal mg. Figure out the correct normal force by analyzing the vertical components of the forces acting on the crate.

The frictional force is equal to your coefficent x the normal force of the box (that is the force exerted by the ground on the box).

The normal force of the box if he's pushing down on it at an angle of 30 degrees will be greater, that is,

R = Fsin(30) + mg

where Fsin(30) is the component of the force in the y direction.

so frictional force = 0.25 * (Fsin(30) + mg)

Does that make sense?

(Apologies for stepping on your toes Doc Al)

## 1. What is the difference between work and kinetic energy?

Work is the transfer of energy that occurs when a force is applied to an object and the object moves in the direction of the force. Kinetic energy is the energy an object possesses due to its motion. In other words, work is the cause of a change in kinetic energy.

## 2. Which one is more important - work or kinetic energy?

Both work and kinetic energy are important concepts in physics, and they are closely related. Work is necessary to change the kinetic energy of an object. Without work, an object would not have any kinetic energy. Therefore, both concepts are equally important.

## 3. Can work and kinetic energy be negative?

Yes, both work and kinetic energy can be negative. Negative work occurs when the force applied to an object is in the opposite direction of the object's motion. This means that the work is taking energy away from the object, resulting in a decrease in kinetic energy.

## 4. How is work and kinetic energy related to each other?

As mentioned earlier, work is the cause of a change in kinetic energy. This means that work and kinetic energy are directly proportional to each other. If the work done on an object increases, its kinetic energy also increases. Similarly, if the work done on an object decreases, its kinetic energy also decreases.

## 5. What are some real-life examples of work and kinetic energy?

Some common examples of work and kinetic energy in everyday life include throwing a ball, riding a bicycle, and pushing a cart. In all of these examples, work is being done to move an object, resulting in a change in its kinetic energy.