A farmer uses a tractor to pull a 150 kg bale of hay up a 15° ...

Question

A farmer uses a tractor to pull a 150 kg bale of hay up a 15° incline to the barn at a steady 5.0 km/h. The coefficient of kinetic friction between the bale and the ramp is 0.45. What is the tractor's power output?

Accepted Answer

Hey, everyone. So this problem is dealing with power. Let's see what they're asking us. We have a motor pulling a 255 kg cart filled with crushed stone up rail bars that are inclined at 30° at a uniform speed of 200 cm/s. The Kinetic friction coefficient between the cart and the rail bars is given to us as 0.25. And they're asking us to determine the power output of the motor. So we can recall first, that power is equal to work over delta T and work in turn is given as force multiplied by the displacement. So that kind of tells us that where we're going, we're going to need to figure out the force of this motor. So I'm going to draw a diagram here. So we can see all of the forces acting on this cart. So we're told that we're on an incline That's 30°. I'll draw the court here. So from here, we know that we have the weight acting in a direction um that is perpendicular to the surface of the ground, not the surface of the, um not the surface of the, the rail bar, our normal force however, is perpendicular to the surface of the rail bars. And then we have the force of the motor, moving it up the bars and then we have our friction force opposite of the direction of motion. So we're going to call this plane or X Y axis beer because most of our forces are, are in that direction. So it just makes the, the map a little bit easier. All right. So from here, we can recall that Newton's second law tells us the sum of the forces is equal to mass times acceleration. And we can look at those forces in both their X component and their Y component. So some of the forces in the direction are going to be have positive F of the motor negative F friction. And then our weight component is going to be weight multiplied by the sign of beta because we are working at a uniform speed. We know that our acceleration is zero and therefore some of these forces is equal to zero. So we have force of our motor is equal to our friction force plus weight times the sign of theta. In turn, we can recall that our friction force is given as the UK and from UK is our friction coefficient and weight is given as mass multiplied by gravity. So from here we have mass gravity is a constant, we have theta and we have our coefficient of friction given in the problem, but we don't have our normal force. So we can't solve for the force of our motor quite yet. And again, we're gonna need that to find work, to find power. So to help find our normal force, we can take the sum of the forces in the Y direction. Again, acceleration is zero. So the, some of those forces is going to equal zero. And in the Y direction, the forces we have are our normal force. And the positive why? And then the Y component of our weight which is given as weight multiplied by cos theta. So our normal force is equal to the wait times the cosine of data. And from there recall that weight is given as mass times gravity, we can plug this quantity in for our normal force. So that looks like the normal force or the sorry, the force from the motor is given as UK multiplied by M G cosine theta plus M G sine beta. We can plug in all of our known values here and solved for the force of the motor. Given in the problem, our coefficient of friction is 0.25 Mass was given as 255 kg. We can recall our constant is our constant gravity is 9.8 m per second squared And the angle given to us is 30. So it's all gonna be multiplied by the cosine of 30. We're gonna add that to our weight, or again, our mass 255 kg multiplied by gravity 9.8 m per second squared Multiplied by the sign of 30. And when we plug that in, we are left with A force from the motor of 1791 Newtons. OK. So I'm going to rewrite the equations that we had written down at the beginning. Power is equal to work over delta T and work is equal to force multiplied by the displacement. The problem here when we put these two equations together, we'll have force multiplied by displacement over delta T, but we don't have our displacement nor do we have our, our change in time. However, we can recall that the change in distance over the change in time is equal to our velocity. That's, that's the definition of velocity. So we can rent this as power equals force multiplied by velocity. And we were given velocity and a problem Of a speed of 200 cm/s. So 200 centimeters per second equals two m per second. We keep that in our standard units and from there, we can plug in Our Force 1791 Newtons Multiplied by our speed, two m/s. And we are left With 3.5, 8 kW. So that is the final answer to this problem. And that aligns with answer choice C so that's all we have for this one. We'll see you in the next video.

A farmer uses a tractor to pull a 150 kg bale of hay up a 15° incline to the barn at a steady 5.0 km/h. The coefficient of kinetic friction between the bale and the ramp is 0.45. What is the tractor's power output?

Key Concepts

Power

Work and Energy

Friction

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