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leva [86]
3 years ago
11

How much work did the movers do (horizontally) pushing a 41.0-kg crate 10.6 m across a rough floor without acceleration, if the

effective coefficient of friction was 0.60?

Physics
1 answer:
VLD [36.1K]3 years ago
5 0

Answer:

The required work done is 2555.448~J

Explanation:

Consider 'F' is the applied force on the crate and 'f' be the force created by friction. According to the figure if '\mu_{k}' be the coefficient of friction, then

f = \mu_{k} \times N = \mu_{k} \times Mg

where 'M', 'N' and 'g' are the mass of the crate, the normal force aced upon the block and the acceleration due to gravity respectively.

Since the application of force by the movers does not create any acceleration to the block, we can write

F = f = \mu_{k} \times M \times g = 0.6 \times 41~Kg~ \times 9.8~m~s^{-2} = 241.08~N

So the work done (W) in moving the crate by a distance s = 10.6 m is

W = F \times s = 241.08~N \times 10.6~m = 2555.448 J

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2. What is the kinetic energy of a 7.26 kg bowling ball that is rolling at a speed of 2 m/s?
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At the bottom, the car's kinetic energy is

                                 KE = (1/2) (mass) (speed²) .

You said that the car's speed is 70 m/s at the bottom of the hill,
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Divide each side by (mass): 

               (0.5) (4900 m²/s²) = (0.9) (9.8 m/s²) (height)

(There goes the mass.  As long as the whole thing is 90% efficient,
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Divide each side by (0.9):

               (0.5/0.9) (4900 m²/s²) = (9.8 m/s²) (height)

Divide each side by (9.8 m/s²):

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                          =  (5 x 4900 m²/s²) / (9 x 9.8 m/s²)

                          =  (24,500 / 88.2)  (m²/s²) / (m/s²)

                          =        277-7/9    meters
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