the work you do when pushing a shopping cart twice as far while applying the same force is This is a topic that many people are looking for. cfcambodge.org is a channel providing useful information about learning, life, digital marketing and online courses …. it will help you have an overview and solid multi-faceted knowledge . Today, cfcambodge.org would like to introduce to you Work Done By a Constant Force and By Friction, Net Work Calculations, Physics Problems. Following along are instructions in the video below:
This physics video tutorial explains how to calculate the work done by a constant force and how to calculate the work done by friction. In addition, it discusses how to calculate the net work done on an object. Whenever the force and displacement vector are parallel to each other, the work done on an object is positive which causes the speed and kinetic energy of the object to increase. If the force and displacement vectors are perpendicular, no work is done and the object’s kinetic energy remains constant. The work done by friction is usually negative since the kinetic friction force vector is antiparallel to an object’s displacement. This video contains plenty of examples and practice problems.
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This video. I’m gonna talk about how to calculate the work done by a a constant force so let’s say if you have a box. And if you a force on the box.
Let’s say in the x direction. And its displacement is along the x direction is the work done by this force is it positive or negative whenever. The force and displacement are in the same direction.
The work done by that force is positive now. Let’s say if we have a box. Where it’s moving towards the left.
But we apply a force on the right what’s gonna happen the work done by this force is now negative because the force and displacement vectors are anti parallel. The force is positive because it’s in the positive x direction. But the displacement is negative.
So if you multiply a positive by a negative value you will get a negative result. Now when the work done by a force is positive the kinetic energy of the object increases the object speeds up in the second example. The object is moving to the left.
But it fills an acceleration towards the right force and acceleration have the same direction. And so if you’re moving to the left. But accelerate towards the right then the object is slowing down.
So therefore the kinetic energy decreases. So anytime work is positive the object speeds up and when it’s negative the object. It slows down the speed decreases now let’s say this is the ground level and there’s a ball and the ball is fallen due to the force of gravity that is acting on the ball.
So it’s also moving in this direction it has the displacement at a negative. Y direction. So the work done on the ball is it positive or is it negative now the force of gravity is in the negative y direction so it’s negative and the displacement is in a negative y direction work is equal to force times displacement especially when these two are parallel so we have two negative values therefore we should expect that the work done on the ball is positive now whenever the work done on an object is positive what’s gonna happen to the speed.
Any time the work done is positive the speed increases and if you drop a ball from the top of the building. It’s gonna fall down faster and faster and faster. If you ignore the effects of air resistance the speed will continue to increase now the velocity is negative.
And it’s becoming more negative. But speed is always positive so the speed goes up which means the kinetic energy of the ball increases. So we can say the work done on the ball is positive.
Because it increases the kinetic energy of the ball. So we see that an object that has fallen towards earth the work done by the gravitational force on the object is positive because the object is speeding up towards the earth. But let’s say if we throw an object up so it’s moving in a positive.
Y direction. But it still has the force of gravity and a negative y direction. What happens so when you throw a ball upward while it’s moving upward against the force of gravity.
The work done has to be negative the work done on the ball. The force of gravity is still negative. But the displacement.
So if you multiply these two you get a negative you zone and while the ball is going up the speed of the ball decreases and so the kinetic energy is decreasing. While it’s going up and then eventually the ball is going to reach a maximum height and then it’s going to begin to fall. And when it falls.
The work done by gravity is positive now since force and displacement are in the same direction. And the object begins to speed up as it falls back to the ground and so the kinetic energy is increasing. I’m going to falls back down.
Now let’s move on to circular motion. So let’s say this is the earth and this is let’s say that’s the moon and so. The moon is in orbit around the earth.
Now the earth exerts a gravitational force on the moon that accelerates. It towards the earth. So does.
This force exert any work on the moon or does it do any work on the moon. This woman say so the force of gravity on the moon does it do any work on the moon. What would you say now the moon is moving in this direction.
It has a displacement in a positive. Y direction and a force in the negative x direction. So the combination of these two the moon is going to turn instead of going straight up or towards the so.
The moon remains in orbit around the earth. Now because the force and displacement vector are perpendicular. No work is done the work.
Done is zero in fact work is equal to force times displacement times. The cosine of the angle between them so the angle theta is 90 and cosine 90 is equal to zero. So whenever.
The force and displacement vectors are perpendicular. No work is done so since gravity does no work on the moon. The moon is gonna remain in orbit now.
If gravity was doing work on the moon. The moon would eventually accelerate towards the earth for any satellite. That falls towards the earth the work done on a satellite is positive as it falls towards the earth.
The speed will increase and so it’s kinetic energy will go up now if a satellite is moving fast enough where it can escape. There is gravity as it moves away from the earth. It’s going to slow down.
And so the work done will be negative. Because the speed is decreasing now to illustrate this let’s draw it on a horizontal surface so i can compare everything so. Let’s say.
This is the ground. Which is the surface of the earth. Now let’s say if we have a ball if the ball falls towards the earth.
We said the work done is positive we have a force of gravity that is directed in the negative y direction. And if the displacement is in the same direction then the work done is positive now the second thing that the ball can do is instead of falling down. It can go straight.
This is basically what the moon is doing the moon is not falling towards the earth. And it’s not moving away from the earth. It’s maintain its orbit.
So we have a force that directs it towards the earth. And it’s moving in the x direction. So the moon is gonna follow the curvature of the earth.
But whenever force and displacement perpendicular. No work is done so the object continues to move at constant speed. Here.
The speed is increasing and so. There’s an increase in kinetic energy. Here.
The change in kinetic energy. Is the same because the speed is constant. Now if we have a ball that moves away from the earth.
It’s basically it’s going up the work. Done is negative. You have an upward displacement and a downward force.
And so as it moves away from the earth. It’s gonna slow down. So this is equivalent to throwing the ball up when you throw a ball up it slows down until it reaches its maximum height.
So the work done for any object that’s moving away from the earth is negative. But if it moves towards the earth is positive if it stays leveled or in orbit around the earth. The work.
Done is zero. Now let’s focus on this question. How much work is done by a horizontal force of 150 newton’s that acts on a crate for a distance of 10 meters.
So let’s draw a picture. So let’s say if we have a crate and we’re applying a horizontal force of 150 newton’s and let’s say the crate moves with a displacement of positive 10 meters towards the right now because the force and displacement vectors are in the same direction. The work done by this force is going to be positive.
So worth is going to be forced times displacement and multiplied by cosine theta so between these two vectors the angle between them is zero. So we have the force of 150 the displacement of 10 times cosine of zero degrees. So.
Cosine. 0 is 1 so it’s 150 times. 10 so it’s 1500.
Joules that’s the unit for. Work so the work done by the force is 1500. Joules.
Now what about part b. How much work is done on the crate across a frictionless surface if there’s no. Friction the work done on the crate is gonna be 1500.
Joules. Because the crate is going to receive the 1500 joules that was transferred to it by a force a force is capable of transferring energy to an object so when we say how much work is done by the. Force it’s.
1500. Joules now the work done on a crate is also 1500. Because it receives it if there’s no friction.
The crate will begin to speed up and so the kinetic energy of the crate will increase now what about part c. How much work is done on a crate if the crate returns to rest due to friction. So imagine push in a box across the surface.
And then eventually the box comes to us how much work did you do on a crate or on a box in this case. Then that work is zero because initially you increase the speed. But if it’s speed.
We turn back to zero its kinetic energy is zero. So there was no change in kinetic energy. So for part c.
The work done on a crate is going to be zero. And let me explain why so we have a force of 150 newtons. So the work done by that force is 1500 joules.
Now if the crate goes to rest that means. There’s a frictional force that’s acting on it and so the energy that was transferred by the force was eventually transferred to friction because friction brought it back down to rest. So the work done by the frictional force is negative.
Fifteen hundred joules. It’s negative. Because the frictional force is opposite to the direction of the displacement vector and also friction tends to slow things down.
It tends to decrease the kinetic energy of an object as opposed to increase in it so the net work done on a crate is the sum of the work done by the force. Plus. The work done by the frictional force so it’s going to be positive.
Fifteen hundred plus. The negative fifteen hundred so the work done the net work done on a crate is zero all of the energy. That was transferred by the force was eventually transferred to friction.
Which was lost due to the thermal energy. Friction generates heat. It generates thermal energy.
So all of the energy that the force exerted or that the force transfer to was eventually transferred to friction. Which eventually was converted to heat. So there was no net work done on the crate and the net work done on the object is equal to the change in the kinetic energy of the object.
So if the object returns to rest its initial speed is the same as the final speed at zero. Which means. The change in kinetic energy is zero.
So then that work done is going to be zero. So if you need to determine the sign of work if you want to determine if it’s positive negative or zero. There’s two ways to do it first compare the force and the displacement vector if the force and displacement vector on the same direction the work done is positive if they’re in opposite directions.
The work done is negative. And if they’re perpendicular no work is done the secondly to determine the sign of work is to look at how it affects the kinetic energy of the object. Does it increase the speed or decrease the speed so if the kinetic energy increases.
The speed increases so any force that increases the speed of an object is doing positive work on that object and any force that is decrease in the speed. Where the kinetic energy of the object is to a negative work on the object now any force that does not affect the speed. So if the change in kinetic energy is zero.
I forgot the equal sign. Then the speed is constant. So that force is doing no work on the object.
So that’s the second way to determine the sign of w. Number to a forty kilogram box is pulled 100 meters by a tension force of 200 newtons. At 30 degrees above the horizontal to the right.
A constant kinetic frictional force of 80 newtons impedes. The motion of the box. What is the work done by the tension force so let’s draw a picture.
So we have a 40 kilogram box. And it’s being pulled to the right by a tension force and that force is 30 degrees above the horizontal. Now this tension force is 200 newtons and is a constant frictional force that slows it down.
And it’s 8 newtons. What is the work done by the tension force now the tension force has an x component. Which we can call tx and tx.
It’s parallel to the displacement of the box. The box is moving to the right so the work done by the tension force is equal to the x component of the tension force times the displacement. Because those two are parallel the angle between them is zero and cosine zero.
Is one. Now tx is well t. Cosine theta.
Now the angle between t. And d. Is not sir.
That angle. Is 30. So thus.
We have the equation t. D. Cosine.
Theta. Or fd cosine theta. So let’s go ahead.
And calculate it the tension force is 200. The displacement is 100 meters and we have an angle of 30 degrees. So it’s 200 times 100.
Which is 20000 times cosine 30. So i got seventeen thousand three hundred twenty point five duels. So that’s how much work.
The tension force does on the block or the box. Now the work is positive because tx and d are both in the positive x. Direction.
They’re both moving towards the right so positive work is being done by that force. And i’m just gonna be write the answer here so i can focus on part b. So.
Let me go ahead and erase this now. What is the work done by friction. The work done by friction is going to be the frictional force multiplied by the displacement.
And then you could put cosine theta. If you want as well. Now.
The frictional force has a magnitude of 80 newtons. The displacement is 100 meters and now the angle between these two what is it now let’s say if we have a force and displacement vector the angle between these two is zero degrees. Now what if the displacement is to the right.
But if the force is to the left. So what we have is a picture that looks like this the angle of a straight line is 180. So because those two arrows are anti parallel one points to the right will influence the left the angle is 180.
So this is cosine 180 cosine. 180. Is negative one.
So this becomes a negative 80 times 100. So. If you didn’t want to bother with the angle.
You could just say that this force is going in the negative x direction. So therefore we can say it’s a negative. 80.
And the answer will be the same so negative 80 times 100 is negative 8000. So that is the work done by friction. So what is the net work the net work done on the crate or in a box.
It’s going to be the sum of the work done by the tension force plus. The work done by friction and so what we need to do is add these two values so it’s seventeen thousand three twenty point five plus negative eight thousand so then that work is nine thousand three hundred twenty point five joules now the other way in which to calculate the net work is to use the net force equation and we should get the same answer if we do it so then that work is equal to the net force times. The displacement.
Now tx is a force that’s directed towards the right. And then we have the frictional force that impedes it so in the x direction. The net force is equal to tx minus the frictional force and so we have tx.
Which is t cosine theta minus. The frictional force so the fictional force carries a negative sign because it’s towards the left. So that’s why it’s minus.
The frictional force. So then that work should be 200 cosine. 30.
Minus. The frictional force of 80 times. 100 so 200.
Cosine. 30 that’s 170 32. That’s the horizontal force of the tension force.
I should say that’s the horizontal component of the tension force that’s what i meant to say and so when he’s is attracted by. 80 the frictional. Force so the net force in the x direction is.
932. Newtons. And then if we multiply that by a hundred that’s going to give us the net work that’s done on the block.
Which is nine thousand three hundred twenty point five. So that’s another way in which you can calculate then that work done on an object just multiply than that forced by the displacement. If the net force and the displacement are parallel if they’re not then you need to take into account the angle.
That’s between them so anytime. You have a force and displacement. That are parallel you could say the work is simply f.
Times. D. But if they’re not parallel.
Let’s say if they’re at an angle. Then the work is now fd cosine theta where d is the displacement. .
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