A factory worker pushes a 30.0-kg crate a distance of 4.5 m along...

Question

A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and the floor is 0.25.(b) How much work is done on the crate by this force?

Accepted Answer

Hey everyone today, we're being asked to calculate the work done on a desk by a force in a horizontal direction. So let's go ahead and look through the question, find some important details and then go ahead and think about theoretical and then finally solve the problem. So we're told that there is a constant horizontal force F that is applied to a 60 kg desk. That is our mass. It is moved six m horizontally and has no acceleration. The coefficient of kinetic friction between the desk and the floor is also 0.5. So with that in mind, let's go ahead and think about this a bit and let me draw this out for you. So, if we have a surface, a horizontal surface, as it says in the problem, we have a box. This is our desk, let's say. So the desk itself, oops! The desk itself is 60 kg which means it has a weight of MG, which is equal to kilograms. Multiplied by the acceleration due to gravity or 9.8 m per second squared. So that is the weight. We're being asked to use kinetic friction as well because we know we're at a constant horizontal force and there's zero acceleration, which means we have a constant velocity. So, if there's no acceleration, that means that the force being applied, the force applied must be equal to the force of kinetic friction. And we know that the force of kinetic friction is equal to the coefficient of kinetic friction multiplied by the normal force. And what is the normal force exactly. Well, it is the force that is perpendicular to the surface of travel. And in this case that will be directly up. And since it will be opposite because it has to repel or essentially prevent whatever object we have from going into the ground. The normal force will also be equal to MG. So we can rewrite the force of kinetic friction as the coefficient of kinetic friction multiplied by MG. So that has to do with our force. That is our total force working here and if we consider the definition of work will work is equal to the force, multiplied by distance, multiplied by cosine Theta. And in this case there is no angle between the object travel and the horizontal it is traveling in direction with the horizontal, this is the direction. So theta is equal to zero, which means since its FD cosine zero, Cosine zero is just equal to one. So we get that work is simply equal to the force times the distance. And substituting in our values, we know that force is equal to the force due to kinetic friction multiplied by MG. Multiplied by the distance substituting in our values. Now we get 0.5 is our coefficient of kinetic friction. Our mass is 60 kg, 60 kg acceleration due to gravity as mentioned earlier is 9.8 m per second squared and the distance traveled is six m is six m, calculating all this and solving we get a final answer Of 1000, 764 jewels or answer choice c. Therefore, the work done on the desk by the force f factoring in friction is 1,764 jules. I hope this helps, and I look forward to seeing you all in the next one.

A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and the floor is 0.25.(b) How much work is done on the crate by this force?

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