A box moves 5\,\text m5m5, start text, m, end text horizontally when force F=20\,\text NF=20NF, equals, 20, start text, N, end text is applied at angle \theta=45\degreeθ=45°theta, equals, 45, degree. What is the work done on the box by FFF during the displacement?

70.71

Explanation:

Given that :

Force (F) = 20N

Distance moved (d) = 5 m

θ = 45°

Workdone on the box :

Workdone (W) = Fdcosθ ; where ;

F = Force, d = Displacement, θ = angle

W = 20 * 5 * cos45°

W = 100 * 0.7071067

W = 70.71 Nm

explanation:

newton's third law, for every action there is an equal and opposite reaction.

the correct choice is a .

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