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A 500N box rests on a horizontal floor. A constant horizonta...
physics waec-physics-2020 ssce-exam-2020 waec-2020
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A 500N box rests on a horizontal floor. A constant horizontal force is exerted on the box so that it moves through 8m. If the coefficient of kinetic friction between the floor and the box is 0.22, calculate the work done on the box.

A. 880J

B. 440J

C. 400J

D. 110J

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peter
Asked by peterrep:1.6K
University of Lagos Nigeria
29 August 2020

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james_mark
James_mark answeredrep:694
School not set Nigeria
23 October 2020
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Fr is the force applied on the box which causes it to move.

F is the force exerted by th box

μ is the coefficient of friction.

Fr = 0.22 x 500

= 110N

Therefore, the work done, W = Fr x d

where,

Fr is the force applied on the box which causes it to move and d is the distance moved

W = 110 x 8

= 880J

Answer: A

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