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Work, Power and Energy

Question : 1
Total: 30
Work done W = F⋅dcosθ1where θ is the angle between the direction of force vector
and displacement vector
(a) Work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket is positive because lifting a bucket out of a well, force equal to the weight of the bucket has to be applied by the man along the vertical in upward direction. Since displacement is also in upward direction θ = 0°. Therefore
W = F⋅dcosθ = F⋅dcos0° = Fd. (positive).
(b) Work done by gravitational force in the above case is negative because the bucket moves in a direction opposite to the gravitational force which is acting vertically downwards. The angle between the gravitational force and the displacement is 180°. Therefore,
W = F⋅dcosθ = F⋅dcos180° = – F⋅d. (negative).
(c) Work done by friction on a body sliding down an inclined plane is negative. Friction always acts in a direction opposite to the direction of motion. Therefore
θ = 180°, W = F⋅dcosθ = F⋅dcos180° = –Fd. (negative).
(d) Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity is positive. Because applied force and the displacement are in same direction. Therefore
θ = 0°, W = Fdcosθ = Fdcos0° = Fd (positive).
(e) Work done by the resistive force of air on a vibrating pendulum in bringing it to rest is negative because direction of resistive force is opposite to the direction of motion of the pendulum. Therefore
θ = 180°, W = Fdcosθ = F⋅dcos180° = –Fd (negative).
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