A gardener pushes a 12 kg lawnmower whose handle is tilted up 37° above horizontal. The lawnmower's coefficient of rolling friction is 0.15. How much power does the gardener have to supply to push the lawnmower at a constant speed of 1.2 m/s? Assume his push is parallel to the handle.

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Hey, everyone. This problem is dealing with friction forces. Let's see what they're giving us and what they're asking from us. We have a road worker moving A kg Machine by pushing it with a force of magnitude f making an angle of 25° below the horizontal. The machine rolls at a constant speed of half a meter per second on a horizontal surface. They also give us the coefficient of friction and they're asking us to find the power delivered by the worker. So the first thing we can do here is recall that P power is given by F V times cosine of theta. So I'm gonna draw a little diagram here so that we can look at all the forces acting on this machine. So we have the machine, we know we have the friction force, it's going to be working against the direction of motion. We've got the weight of the machine, the normal force and then we have the force of the worker And we know that that is going to just draw another horizontal line here. And that angle is 25°25 below the horizontal. And so we'll call that the pushing force or F P. So to solve for power, we do have speed from the problem that we don't have force. So we're gonna need to solve for the force first and then we can solve for power. So if we can recall that the speed is, if we know from the problem that the speed is a constant speed, that means our acceleration is zero. So we can use Newton's second law of F equals M A where acceleration is zero to be zero. So the sum of the forces are going to be zero as well. So we'll rewrite that as some of the forces in the X direction or zero And the sum of the forces in the Y Direction are zero. So let's take The sum of the forces in the Y Direction 1st. So we know that in the Y direction, we have the normal force in the positive Y, we have weight in the negative Y and then we have F P and then the Y component of that is going to be a sign of data. And because it's below the horizontal, that's also a negative And that all equals zero. So our normal force, our weight is mass times gravity. And then we have F P times sine of theta in our X direction, we have the friction force in the negative. And then in the positive X direction, we have F P Times Cosine of Theta zero. So we can write this as FP cosign theta equals F F. We know that our friction force is given by U K or coefficient of friction times and our normal force. And so we've got these two equations, we have a number of unknowns. We don't know N or a normal force and we don't know F P, but we can use these equations, we can plug one into the other to solve for S P. And we're going because both of those equations have this normal force. So that's the next step here. So we'll solve this first equation for N. So that's going to look like N equals F P times the sine of theta plus MG. And we will plug that over in over here for this. And so now we have F P co sign theta equals U K times FP sign data plus. Okay. So we're just gonna keep simplifying this until we get to a point where we can solve for F P. So we do know from the problem that the cosine of theta is 25 And UK was .18. Let's just write that here. Data is 25° UK is 0.18 and our weight It's 32 kg Or mass, sorry, our mass is 32 kilograms. And so from here, we can take F P times cosign of theta minus the sine of theta. No sign of data over um UK is equal to MG. And then from there, we'll plug in our known values. So co sign of 25 over 250.18 minus sign of 25 equals 32 kg Times gravity a constant we can recall is 9. m/s squared. You can solve that out for F P and we'll get 13.9 Over 4.61 which equals 68. new. And then we're gonna go back up to that very first equation that we were called for power. So P equals F V times cosine theta. And so the force that we just found 68.1 newtons times the velocity given to us in the problem 0.5 m per second times the cosine of Equals 31 Watts. So that is the answer to the problem. We look at our potential solutions and that aligns with choice B. That's all we have for this one. We'll see you in the next video.

A gardener pushes a 12 kg lawnmower whose handle is tilted up 37° above horizontal. The lawnmower's coefficient of rolling friction is 0.15. How much power does the gardener have to supply to push the lawnmower at a constant speed of 1.2 m/s? Assume his push is parallel to the handle.

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Key Concepts

Forces and Motion

Friction

Power