Question

2.8. A body moves a distance of 5 m on a smooth horizontal surface under the influence of a force of 20n
calculate the work done by the force when
(a) the force acts along the horizontal surface and
(b) the force acts along a direction at an angle of 60' with the horizontal surface.​

Answer 1

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Work done is given by F.S(dot product of force and displacement).

(a).When body moves along inclined plane force required is mgSinθ.

=When body moves vertically upwards then force required is mg.

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(b).When body moves horizontally force required is equal to the frictional force which is μ.mg where μ is generally between 1 and 0 .

So maximum force is required when moving vertically upwards and displacement in each case is same.

When pushed over smooth roller then force required is almost zero. So work done is zero.

So maximum work done is in case of vertically upwards.

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Answer 2

Answer: The work done in (a) is $100 J$ and in (b) is $50 J$.

Explanation:

Given: Distance $= 5m$

Force $= 20 N$

To find: (a)  Force which is acting along the horizontal surface

(b) Force which is acting at an angle of $60$° with the horizontal surface.​

Solution: Work done $(W)$ on a body to move with a force $(F)$ to a certain distance $(S)$ at an angle θ is calculated as:

$W = F S Cos$ θ

(a) In first case, the force is acting along the horizontal surface; ∴ θ $=0$°

$Cos 0$° $= 1$

Work done $=FS$ $= 20 N * 5 m= 100 Nm = 100 J$

(b) In second case, the force is acting at an angle of $60$° with the horizontal surface; θ $= 60$°

As $Cos 60$° $= \frac{1}{2}$

Work done $= FS Cos 60$° $= 20 N * 5 m * \frac{1}{2} = 50J$

The work done in (a) is $100 J$ and in (b) is $50 J$.

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