# How Do You Calculate Paul's Speed Using Work and Energy?

• mawalker
In summary, the conversation involved solving a physics problem involving a baby brother sitting on a mat being pulled by his sister with a constant tension and coefficient of friction. The goal was to find the baby's speed after being pulled a certain distance. The steps involved setting up a force diagram, finding the normal reaction force, and determining the net horizontal force. The net work in the horizontal direction was then used to calculate the baby's speed.
mawalker
I am working on the following problem.

Susan's 13.0 kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30 degrees above the floor. The tension is a constant 31.0 N and the coefficient of friction is 0.190 . Use work and energy to find Paul's speed after being pulled 2.60 m.

So far I have set up my force diagram with my x-axis along the ground. I have normal force pointing up. weight pointing down. tension to the left. and force susan to the right. I'm really not too sure where to go from here though. Any thoughts?

Your tension should not be pointing left, it should be inclined at 30 degrees to the horizontal. Your next step should be to find the normal reaction force.

Hootenanny said:
Your tension should not be pointing left, it should be inclined at 30 degrees to the horizontal. Your next step should be to find the normal reaction force.

thanks hootenanny. i fixed my force diagram and i calculated the normal force to be 13kg * 9.8 ms = 127.4 N. Am I on the right track?

mawalker said:
thanks hootenanny. i fixed my force diagram and i calculated the normal force to be 13kg * 9.8 ms = 127.4 N. Am I on the right track?
Not quite. Look at all the vertical forces acting, you should have three;
1. Weight of the child
2. Normal Reaction Force
3. ?
What's the third?

the force of friction acting along the incline

mawalker said:
the force of friction acting along the incline
Not quite, there is no incline, the friction is acting horizontally. In addition to the two forces outline above, you will also have a component of the tension acting vertically (and a component acting horizontally). Can you see why?

yeah i think so, so would the force of tension vertically just be the 31 N * sin (30 degrees) = 15.5 N added to the 127.4 and i get 142.9 N total force. Is this correct?

mawalker said:
yeah i think so, so would the force of tension vertically just be the 31 N * sin (30 degrees) = 15.5 N added to the 127.4 and i get 142.9 N total force. Is this correct?
Take a careful look at the directions of the vertical forces.

right, the tension force would be pointing vertically downwards so the 15.5 would be subtracted from 127.4 and total force would be 111.9 N?

mawalker said:
right, the tension force would be pointing vertically downwards so the 15.5 would be subtracted from 127.4 and total force would be 111.9 N?
The vertical component of the tension acting where the rope is attaced to the mat is upward, in the same direction as the normal. The normal force is upward, opposing the weight.

T_y + N = mg

So yes, the normal force is the weight minus the vertical tension component.

gotcha...i'm still kinda lost on where to go from this point now

mawalker said:
gotcha...i'm still kinda lost on where to go from this point now
From the normal force you can find the friction, which opposes the horizontal component of the tension in the rope. The resultant of those two is a net horizontal force that accelertes Paul. You can find the acceleration, but the problem asks you to use work and energy to find the speed after a certain distance. What is the net work done by the force in the horizontal direction?

ok so to find the force of friction i took the coefficient of friction .190 times the normal force to give me 21.26 N. The tension in the horizontal direction would be 15.5 N, which would give me 36.7 N as the net work in the horizontal direction.

mawalker said:
ok so to find the force of friction i took the coefficient of friction .190 times the normal force to give me 21.26 N. The tension in the horizontal direction would be 15.5 N, which would give me 36.7 N as the net work in the horizontal direction.
That is not the horizontal component of the tension. That is the vertiacl compoonent. When you do find the horizontal component, remember that friction opposes the motion.

## 1. What is the work-energy theorem and how is it related to force diagrams?

The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy. This means that the amount of work done on an object by a force is directly related to the change in the object's energy. Force diagrams are used to represent the different forces acting on an object, and by analyzing these diagrams, we can determine the work done on the object and its resulting change in energy.

## 2. How do you calculate the work done on an object using the work-energy theorem?

To calculate the work done on an object using the work-energy theorem, you need to first determine the net force acting on the object. Then, multiply the net force by the distance the object moves in the direction of the force. This will give you the work done on the object, which is equal to the change in its kinetic energy.

## 3. Can the work-energy theorem be applied to both linear and rotational motion?

Yes, the work-energy theorem can be applied to both linear and rotational motion. In linear motion, the work done is equal to the force applied multiplied by the displacement of the object. In rotational motion, the work done is equal to the torque (rotational force) applied multiplied by the angular displacement of the object.

## 4. How does the work-energy theorem relate to the conservation of energy?

The work-energy theorem is closely related to the principle of conservation of energy. Conservation of energy states that energy cannot be created or destroyed, only transferred from one form to another. The work done on an object changes its energy, but the total amount of energy in the system remains constant. This means that the work done on an object is equal to the change in its energy, and this change is reflected in the conservation of energy.

## 5. What are some real-life examples of the work-energy theorem and force diagrams?

One example of the work-energy theorem and force diagrams in action is a roller coaster. The force diagram would show the different forces acting on the roller coaster, such as gravity, normal force, and friction. By analyzing the force diagram and applying the work-energy theorem, we can determine the work done on the roller coaster and its resulting change in kinetic energy. Another example is a pendulum, where the force diagram would show the force of gravity and the tension in the string, and the work done on the pendulum would change its gravitational potential energy into kinetic energy and vice versa.

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