# Work & Zero Work in High School Physics Course

• chubbyorphan
In summary: child and sled, the child and sled are both experiencing a force that is directed along the horizontal axis.
chubbyorphan
Hi, I am taking a high school physics course and I just need a little help grasping the concept of zero work and whether or not it applies to this situation... The question reads:

A child on a sled (having a combined mass of 47.0kg) is pulled by a force directed along a rope that make a 45° angle with the horizontal axis. The force exerted on the rope is 100.0N. The force of friction acting on the sled is 30.0 N. If the child is pulled a distance of 10.0 m along a level field, determine the total work done on the child and on the sled.

So I worked this out using W = F(cos θ)• ∆d
I calculated the work from the pull and then the work from friction and added the values to get 407 J which I'm pretty confident is right..

Where I need help is in deciding whether or not this value is the work done on the child and the sled..
OR
is this the work done on the sled. And the work done on the child is zero. I'm at this dilemma because my text says that 'whenever a force is exerted perpendicular to the direction of displacement, it does not contribute to forward motion.'
So I'm thinking that the child is just kind of along for the ride and the sled is experiencing work. I think this because the child is only experiencing ForceNormal and ForceGravity.. but then again isn't the child experiencing some kind of ForceFriction from the sled and does the fact that the child's weight is definitely affecting the answer something to consider? Please help!

What you've worked out is the work done on the combined system of child+sled. And that's perfectly legit. There is no reason why you cannot consider the child and the sled to be a single object having their combined mass, since the child and sled are rigidly attached (more or less). When one moves, so does the other.

What keeps the child and the sled attached to each other? Friction, of course. Friction between the child and the seat. Without this friction, the sled would slide out from underneath the child when pulled on. In fact, since you know that the acceleration of the child is the same as the acceleration of the overall system, you can work out the net force on the child (due to this friction) and hence the work done on the child alone.

chubbyorphan said:
I calculated the work from the pull and then the work from friction and added the values to get 407 J which I'm pretty confident is right.
I think that's right.

Where I need help is in deciding whether or not this value is the work done on the child and the sled..
OR
is this the work done on the sled. And the work done on the child is zero.

The sled and the child are accelerating, because of work being done on the sled. Assuming the child is not moving relative to the sled, the child and sled both have the same change in velocity and they both have mass. So the kinetic energy of both is increasing.

Each kg of mass of the child+sled has the same increase in velocity, so each kg of mass has the same increase of KE. You know the total increase in KE (= the work done on the system) so you can split the total into the work done on the child and the work done on the sled.

If you want to work this out using forces, the sled is applying a force to the child to accelerate it (i.e. the friction force between the child and the sled), and the child is applying an equal and opposite force to the sled. But you can answer the question using energy without finding the magnitude of that force, and that is one reason why "energy methods" are useful in mechanics.

Alright, so far I'm following..
Couple things though.. if I drew a FBD of the child.. in regard to just the horizontal forces..
would the child have both Forceofsled and Forcefriction pulling it forward.. is that right?

also I am confused as far as how to split the forces between the sled and the child..

so far I'm thinking:
I feel like the question doesn't give me enough information. Dont I need to know either the coefficient of static friction or forcenormal? and as far as I can tell I need to know the mass of the child without the sled in order to calculate the child's Forcenormal.

chubbyorphan said:
Alright, so far I'm following..
Couple things though.. if I drew a FBD of the child.. in regard to just the horizontal forces..
would the child have both Forceofsled and Forcefriction pulling it forward.. is that right?

What is "forceofsled"?

Thanks to friction between the sled and the child, the child is accelerating along with the sled. That's it. That's the only horizontal force acting on the child.

chubbyorphan said:
so far I'm thinking:
I feel like the question doesn't give me enough information. Dont I need to know either the coefficient of static friction or forcenormal? and as far as I can tell I need to know the mass of the child without the sled in order to calculate the child's Forcenormal.

Enough information to do what? You've already solved the problem, haven't you? You've found the work done on the combined mass of the child+sled.

In order to find the work done on the child alone, you'd need to know the mass of the child, it's true. If you did know that, then you would be able to compute the work done on the child without computing any of the forces on the child at all, as AlephZero has already pointed out. You can just use energy methods.

sweet baby jesus.. I understand

Thanks so much to the both of you!

one more thing.. part b) asks..determine the childs final speed at the end of 10.0m
does this look right?

W = ∆Ek
W = Ekf - Eki
W = (1/2)mvf^2 – 0 J
407 J = (1/2)(47.0kg) vf^2
vf = √(814/47.0)
vf = 4.16 m/s

chubbyorphan said:
one more thing.. part b) asks..determine the childs final speed at the end of 10.0m
does this look right?

W = ∆Ek
W = Ekf - Eki
W = (1/2)mvf^2 – 0 J
407 J = (1/2)(47.0kg) vf^2
vf = √(814/47.0)
vf = 4.16 m/s

That looks fine to me.

## 1. What is the concept of "work" in high school physics?

The concept of work in high school physics refers to the amount of energy required to move an object a certain distance, against a force. This is typically measured in Joules (J) and is calculated by multiplying the force applied to an object by the distance it moves in the direction of the force.

## 2. How is the concept of work related to the concept of energy?

Work and energy are closely related in physics. Work is the amount of energy transferred to or from an object, while energy is the ability to do work. In other words, work is a measurement of energy, and energy is what allows work to be done.

## 3. What is "zero work" in high school physics?

Zero work in high school physics refers to the situation where no energy is transferred to or from an object. This can occur when there is no force applied to an object, or when the force applied is perpendicular to the direction of motion.

## 4. How is the concept of zero work applied in real-life situations?

In real-life situations, zero work can be applied in various scenarios. For example, if you push a door open with a force perpendicular to the direction of motion, no work is done as the force and displacement are at right angles. Additionally, if you hold an object still, no work is done as there is no displacement. In both cases, no energy is transferred to or from the object.

## 5. What are some examples of work and zero work in everyday life?

Examples of work in everyday life include lifting a book from the ground to a table, pushing a shopping cart, or pedaling a bike. On the other hand, examples of zero work include carrying a book while walking, holding a cup of coffee, or sitting in a chair. In these scenarios, no energy is transferred to or from the object and no work is done.

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