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In this question we have been told that a force of constant magnitude pushes a box up a vertical surface.
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The box moves at a constant speed.
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If the mass of the box is 5 .8 kg and it is pushed 2 .6 m vertically upwards and the coefficient of friction is 0 .35, the angle theta is 30 degrees, we need to, in part a, find out the work done on the box by the force f.
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So first of all we are going to draw a free body diagram of the box and represent all the forces acting on it.
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So first of all we know that the weight is going to act vertically downwards which is going to be mg.
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Now there is also going to be a normal force which will act in this direction from the surface.
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We will call this force n.
01:01
Now the box is moving upwards so the friction force will oppose its motion.
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So the friction force will act in the opposite direction.
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So this is going to be the direction of the frictional force which we will call it f.
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Now if we look at this force f, we can divide it into two components.
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The first component is going to be f.
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The horizontal component is going to be f cos theta.
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And the vertical component is going to be f sin theta.
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So we know that the box is moving at a constant speed which means acceleration is 0.
01:40
So f net is going to be mass into acceleration.
01:44
So f net, the net force on the box is also going to be 0.
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If we write down the equation of equilibrium, first of all in horizontal direction, that is going to be f cos theta is going to be equal to the normal force.
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So n is going to be equal to f cos theta.
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Now sum of forces in vertical direction is also going to be 0.
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So if you look at it, f sin theta is acting upwards minus mg minus the frictional force.
02:27
So we know that the frictional force is equals to coefficient of friction into the normal force.
02:49
In this case, the normal force is f cos theta.
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F cos theta.
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So this is going to be mu k into f cos theta.
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So if we need to find out the force f, f is going to be sin theta minus mu k cos theta equals to mg.
03:21
From this equation, the force f is going to be equal to mg divided by sin theta minus mu k cos theta.
03:31
Now if you put down the or plug in the values, the mass of the box is 5 .8 into the gravitational acceleration which is 9 .8, sin 30 minus the coefficient of kinetic friction is 0 .35 cos 30.
03:55
So the value of f is going to be 288 .7 newton.
04:04
Now we need to find out the work done by this force, work done by the force f which is going to be the force f into d cos theta...