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A 84.7 kg refrigerator rests on the floor. How much work is required to move it at constant speed for 2.25 m along the floor against a friction force of 193.5 N
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24 votes
24 votes
A 84.7 kg refrigerator rests on the floor. How much work is required to move it at constant speed for 2.25 m along the floor against a friction force of 193.5 N

User Ascendants
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1 Answer

12 votes
12 votes

Answer:

Step-by-step explanation:

Formula

F - ff = m * a

W = F * d

Givens

a =0

d = 2.25

m = 84.7

ff = 193.5

Solution

F - ff = ma

F - 193.5 = 0

F = 193.5 N

Notice what this is telling you. The Force applied to the fridge is just enough to overcome friction and not a Newton more

Work = F * d

F = 193.5 N

d = 2.25 Meters

Work = 193.5 * 2.25

Work = 435.4 Joules

User Golam Sorwar
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3.0k points
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