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1/A horizontal force of magnitude 42.7 N pushes a block of mass 3.82 kg across a

ID: 1768061 • Letter: 1

Question

1/A horizontal force of magnitude 42.7 N pushes a block of mass 3.82 kg across a floor where the coefficient of kinetic friction is 0.574. (a) How much work is done by that applied force on the block-floor system when the block slides through a displacement of 3.86 m across the floor? (b) During that displacement, the thermal energy of the block increases by 32.7 J. What is the increase in thermal energy of the floor? (c) What is the increase in the kinetic energy of the block?

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2/

A child whose weight is 261 N slides down a 5.70 m playground slide that makes an angle of 47.0° with the horizontal. The coefficient of kinetic friction between slide and child is 0.0740. (a) How much energy is transferred to thermal energy? (b) If she starts at the top with a speed of 0.341 m/s, what is her speed at the bottom?

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3/

You push a 4.8 kg block against a horizontal spring, compressing the spring by 14 cm. Then you release the block, and the spring sends it sliding across a tabletop. It stops 85 cm from where you released it. The spring constant is 280 N/m. What is the coefficient of kinetic friction between the block and the table?

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4/

In the figure, a 4.4 kg block slides along a track from one level to a higher level after passing through an intermediate valley. The track is frictionless until the block reaches the higher level. There a frictional force stops the block in a distance d. The block's initial speed is v0 = 5.0 m/s, the height difference is h = 1.1 m, and ?k = 0.638. Find d.

Explanation / Answer

4.

Using energy conservation;

0.5mvi^2-mgh=0.5mvf^2

0.5*4.4*5^2-4.4*9.81*1.1=0.5*4.4*vf^2

or

Vf=1.8487m/s

Now using work energy theorem;

0.5mvf^2=Work done by friction

or

umgd=0.5mvf^2

or

d=0.273m

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