A force of 60 N acts on a body which moves it through a distance of 4 m on a horizontal surface. What is the work done, if the direction of the force is at an angle of $60^{\circ}$ to the horizontal surface?

100 J

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120 J

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75 J

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150 J

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Solution

The correct option is B
120 J

Given that F= 60 N
Displacement (s) = 4m
$θ = 60^{\circ}$
Work done = F.S . $c o s θ$
$= 60 \times 4 c o s 60^{\circ} = 240 \times \frac{1}{2}$
Work done = 120 J

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Similar questions

A force of 60 N acts on a body which moves it through a distance of 4 m on a horizontal surface. What is the work done, if the direction of the force is at an angle of $60^{\circ}$ to the horizontal surface?

Q. (i) What happens to the work done when the displacement of a body is at right angle to the direction of force acting on it? Explain your answer.
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Q. (a) What happens to the work done when the dispacement of body is at right angles to the direction of force acting on it? Explain your answer.
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A force of 60 N is pushing a body which moves it through a distance of 4 m on a horizontal surface. What is the work done, if the direction of force is at an angle of $60^{\circ}$ to the horizontal surface?

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A force of 60 N acts on a body which moves it through a distance of 4 m on a horizontal surface. What is the work done, if the direction of the force is at an angle of 60∘ to the horizontal surface?

The correct option is B 120 J Given that F= 60 N Displacement (s) = 4m θ=60∘ Work done = F.S . cosθ =60×4 cos60∘=240×12 Work done = 120 J

A force of 60 N acts on a body which moves it through a distance of 4 m on a horizontal surface. What is the work done, if the direction of the force is at an angle of $60^{\circ}$ to the horizontal surface?

The correct option is B
120 J

Given that F= 60 N
Displacement (s) = 4m
$θ = 60^{\circ}$
Work done = F.S . $c o s θ$
$= 60 \times 4 c o s 60^{\circ} = 240 \times \frac{1}{2}$
Work done = 120 J