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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120 N cm

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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.
(ii) A force of 50 N acts on a body and moves it a distance of 4 m on a horizontal surface. Calculate the work done if the direction of force is at an angle of 60° to the horizontal surface.

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.
(b) A force of 50 N acts on a body and moves it a distance of 4 m on horizontal surface. Calculate the work done if the direction of force is at an angle of 60° to the horizontal surface.

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?

Q. A force of

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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