A dockworker applies a constant horizontal force of 80.0 N to a b...

Question

A dockworker applies a constant horizontal force of 80.0 N to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves 11.0 m in 5.00 s. (a) What is the mass of the block of ice?

Accepted Answer

Hey everyone. So today we're dealing with some kinda matics. So we're being told that a worker is pushing a sled using constant horizontal force with a magnitude of 80 newtons over a smooth ice surface. So this means that we can also neglect a friction just due to ice being very low and friction for the sake of this purpose as well as known in the problem, friction force is negligible. So from rest the sled covers 8.7 m and 2. seconds. With this information, we're being asked to determine the mass of the sled. So as always when we're dealing with cinematics, our first step is to draw out a hey um a force body diagram, right? So we have we have a horizontal icy plane with negligible friction. We have a sled, right? We have a sled that is being pushed so the direction of the motion is here. Or rather the forces being applied. This way It is a force equal to Newtons. We have a weight force which is equal to MG. We have a normal force which is also equal to MG. And we have no friction. So these are all the forces that are acting here, right? And our normal force and weight will cancel out. And this makes sense because when you get to step two, which is using Newton's second law, which states that the sum of all the forces will be equal to mass times acceleration. We know that this means it is 80 newtons is equal to M. A. And while we still don't know the acceleration, we can rearrange just to solve for the mass, which is what we're looking for. And we get that 80 newtons over a will be equal to our mass. So from here we need to figure out our uh mass right? And to do so we need to figure out the acceleration and we can do this using some uniformly accelerated motion equations. Now let us take the distance travel, which will say is delta X as well as the time of the travel. And let us put this into a Kid a matics form. So to say. So we know that the change in position, the Delta X is 8.70 m. The time of the travel was 2.60 seconds. The initial velocity because we started from rest 0m/s. And we don't know the final velocity and the acceleration is our target. So since we know three of the five uh parameters, we can select one of the uniformly accelerated motion equations that doesn't have the unknown, that does not have this uh final velocity in it. So the equation that we can choose and let's read this in blue would be one using the delta X stating that the initial velocity multiplied by time plus half into acceleration into times square will be equal to the change in position. So substituting in our values, we have 8.70 m is equal to zero because it's zero times 2.6 which is still zero plus half into a which we still don't know yet. Oops Still don't know yet. Multiplied by time squared which is 2. seconds squared. It's 2.6 seconds to the power of two and re arranging for a. Oops, we get that A is equal to Is equal to 8.70 m into two, divided by 2.6 square. Just give us an answer of 2. m/s squared. So that is a. Which means going back and solving for mass, which let me do this in red, which means mass will be equal to 80 newtons, 80 Newtons of force divided by an acceleration of 2.57 m per second squared, which gives us a final mass of 31.1 kg. Or answer choice. D Therefore the mass of the sled is 31.1 kg. I hope this helps. And I look forward to seeing you all in the next one

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