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box 5) A 5kg sliding with a speed of 10m/s goes into a rough patch with a coefficient of friction of 0.15. The box slowed down to a speed of 4m/s. How long did it slide in that rough section? What was the work done by the friction?

          box 5) A 5kg sliding with a speed of 10m/s goes into a rough patch with a coefficient of friction of 0.15. The box slowed down to a speed of 4m/s. How long did it slide in that rough section? What was the work done by the friction?
        
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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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box 5) A 5kg sliding with a speed of 10m/s goes into a rough patch with a coefficient of friction of 0.15. The box slowed down to a speed of 4m/s. How long did it slide in that rough section? What was the work done by the friction?
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Transcript

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00:01 In this question we have been told that a box weighting 500 n is pushed across a horizontal floor with a force of 250 n.
00:08 The coefficient of friction is 0 .4.
00:11 In party we need to find out the work done in pushing the box 20 meters and in second part we need to find out how much work went into overcoming friction and how much into accelerating the box.
00:22 So if you look at the free body diagram a 250 n force is applied on the box.
00:28 Its weight is 500 n which is acting downwards.
00:32 Now we know the unknown force is the frictional force is going to act in the opposite direction.
00:38 This is going to be frictional force and a normal force will act upwards to balance out the weight.
00:44 So if we know that the box will move in horizontal direction in vertical direction forces will be balanced which means the normal force is going to be equal to the weight which is 500 n.
01:00 Now we need to find out the work done.
01:05 The total work done.
01:07 So the formula for total work done is f the force applied to the box which is 250 n into d the distance covered.
01:16 So 250 multiplied by the distance covered which is 20 meter.
01:21 So that is 5000 joules.
01:26 That is the total work done and answer for part a...
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