Question

A dockworker applies a constant horizontal force of 80.0 $\mathrm{N}$ to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves 11.0 $\mathrm{m}$ in 5.00 $\mathrm{s}$ . (a) What is the mass of the block of ice? (b) If the worker stops pushing at the end of 5.00 $\mathrm{s}$ s, how far does the block move in the next 5.00 $\mathrm{s?}$

          A dockworker applies a constant horizontal force of 80.0 $\mathrm{N}$ to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves 11.0 $\mathrm{m}$ in 5.00 $\mathrm{s}$ . (a) What is the mass of the block of ice? (b) If the worker stops pushing at the end of 5.00 $\mathrm{s}$ s, how far does the block move in the next 5.00 $\mathrm{s?}$
        
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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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A dockworker applies a constant horizontal force of 80.0 $\mathrm{N}$ to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves 11.0 $\mathrm{m}$ in 5.00 $\mathrm{s}$ . (a) What is the mass of the block of ice? (b) If the worker stops pushing at the end of 5.00 $\mathrm{s}$ s, how far does the block move in the next 5.00 $\mathrm{s?}$
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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? (b) If the worker stops pushing at the end of 5.00 s, how far does the block move in the next 5.00 s?

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Transcript

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00:01 A force of 80 newtons is applied to a block of ice on a smooth, frictionless floor.
00:07 We're told that it starts from rest and is pushed a distance of 11 meters in a time of 5 seconds.
00:17 And we want to know, first of all, what is the mass of this block? and so we're going to first find acceleration.
00:24 And the reason for that is force is equal to mass times acceleration.
00:28 We don't have acceleration, but we have enough information.
00:30 Information to find it.
00:32 And so we're going to use the equation displacement is equal to initial velocity times time plus one -half acceleration times time squared.
00:41 Initial velocity is simply zero.
00:45 And so now we have this equation.
00:47 Multiply both sides by two in the attempt to get acceleration by itself and then divide by t squared or time squared.
00:55 And that gives us our equation for acceleration.
00:58 And so acceleration is equal to two times our distance divided by our time squared.
01:06 And when we do that, we get 0 .88 meters per second squared.
01:12 So now let's plug this into our equation.
01:14 So we have force is equal to mass times acceleration.
01:17 Let's go ahead and divide both sides by acceleration to get mass by itself.
01:22 And so mass is equal to force divided by acceleration.
01:28 And when we plug this in, we get a mass of 90 .9 kilograms to keep it in our three significant figures.
01:36 The next thing we want to know is if at this point the worker quits pushing, he just lets it go...
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