Question

3 A block of mass 10kg is slowly sliding up on a smooth incline of inclination 37° by a person. Calculate
the work done by the person in moving the block through a distace of 2m' if the driving force is applied
a paralel to the incline
b) in the horizontal direction
1) 1001-800
2) -70J, 120J
3) 120, 1200
4)-804,-60J​

Answer 1

Thus the work one when the force along the parallel direction is 120 J and in the horizontal direction is 60 N.

Explanation:

Correct statement:

a block of weight 100 N is slowly slid up on a smooth incline of inclination 370 by as person. Calculate the work done by the person in moving the block through a distance of 2.0 m, if the driving force is a. parallel to the incline and b. in the horizontal direction.

(a) Parallel to the incline

So, work done when the force is parallel to incline,

W = F.d cosθ

= 60 × 2 × cos 0°

60 x 2 x 1 = 120 J

(b) in the horizontal direction.

Mass "m" = 100 N

Angle of inclination "θ" =37°

Distance "s"= 2 m  

Force"F" = mg sin(37°)

= 100 ×3 / 5 = 60 N

Thus the work one when the force along the parallel direction is 120 J and

in the horizontal direction is 60 N.

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Answer 2

a) paralel to the incline- 1200 b) in the horizontal direction- 120

Explanation:

Parallel to the incline :-

When block slides up

driving force become parallel to incline = resisting force of component of weight

given :-

w = 100 N

θ = 37°

distance s = 2.0 m

Driving force F = 100 × sin37° N

Completed Work = F × d

Completed Work = 100 × sin37° × 2.0 m

=1200 J

Hence the work done in the parallel incline is 120 J.

(b) Force in the horizontal direction.

Force of driving happens in the horizontal direction , by incline of its component which is equivalent to the force F which is 100 × sin37° N

So in this case also,

The work done = 120 J. 

Therefore the work done in the horizontal direction is 120 J.

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