An 8.0 kg crate is pulled 5.0 m up a 30° incline by a rope angled 18 ° above the incline. The tension in the rope is 120 N, and the crate's coefficient of kinetic friction on the incline is 0.25.
(b) What is the increase in thermal energy of the crate and incline?

Question

An 8.0 kg crate is pulled 5.0 m up a 30° incline by a rope angled 18 ° above the incline. The tension in the rope is 120 N, and the crate's coefficient of kinetic friction on the incline is 0.25. 
(b) What is the increase in thermal energy of the crate and incline?

Accepted Answer

Hey, everyone. So this problem is dealing with friction forces. Let's see what they are giving us and what they're asking from us. We have an airplane palate of a given mass pushed with a constant force of a given magnitude up a ramp going to start drawing this here. The Applied Force forms a 10° angle above the ramp. So we'll show that here, applied force 10 degrees And the ramp makes an angle of 25° with the horizontal. The coefficient of kinetic friction between the palette and the ramp is given here is .35 and were asked to calculate the change in internal energy of the pallet ramp system due to friction. So first, let's look at all the forces acting on this palette and we're going to define our coordinate system here by the palette. So the angle of the palette here, the palette going to be going in the x direction and then perpendicular to that will be the y axis. So we have our applied force already noted here, you'll also have weight going down. Um We have our normal force perpendicular to the surface and then the friction force is going to be opposite the direction of movement. So from here, we can recall that the change in internal energy is given by delta E, internal equals U K and delta X. So we have U K, that's our coefficient of kinetic friction. And we have delta X or change in distance both given to us in the problem. But we don't have end the normal force. If we look at our force diagram that we drew though, we can see that if we use Newton's second law to some, our forces we can solve for N. So that's our next step here. So the sum of the forces in our wide direction are zero, right? We're not levitating off of the palette. So we can write that as the summit of F Y equals zero. And then we'll look at all of these forces in the Y direction. So we have N which is positive, we have a weight component. Um It's not 100% down or 100% of the, of the weight force. So we need to look at that as MG times the cosine of theta and we have two different angles here. So we're gonna denote that as data, one and data to and then are applied forces the same where it's not the full magnitude of the force. It's going to be the applied force times the sine of theta two equals zero. And so from here, we can plug in what we know to solve for M so N equals so the mass was given to us as 35 kg Times cosign of minus FA was also given to us as 320 Newtons Times The sine of data, two or 10°, plug that into our calculators. And we come up with A Normal Force of 255.6 News. So now we can go back to that first equation and we have everything that we need to solve for the change in internal energy. So μ K was given to us as 0.35, We just sold for N 255.6. And Delta K was given to us in the problem as three m like that into our calculators and get a change in internal energy of 268 jewels. And so that is the answer to this problem. We look at our choices and that aligns with choice b. That's all we have for this one. We'll see you in the next video.

An 8.0 kg crate is pulled 5.0 m up a 30° incline by a rope angled 18 ° above the incline. The tension in the rope is 120 N, and the crate's coefficient of kinetic friction on the incline is 0.25. (b) What is the increase in thermal energy of the crate and incline?

Verified Solution

Key Concepts

Friction and Thermal Energy

Work-Energy Principle

Inclined Plane Dynamics