Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
We have a situation in which, work is done by force acting obliquely.
Force, (F) = 50 N
Distance travelled by the body, (S) = 4 m
Angle between force and direction of distance, (θ) = 60°
We have a relation as,
W = (F)(S) cos θ
So,
`W = (50) (4) cos 60^circ`
= `200 (1/2)` J
= 100 J
So , the work done is 100 J .
APPEARS IN
संबंधित प्रश्न
A force of 7 N acts on an object. The displacement is, say 8 m, in the direction of the force (see the given figure). Let us take it that the force acts on the object through the displacement. What is the work done in this case?

A pair of bullocks exerts a force of 140 N on a plough. The field being ploughed is 15 m long. How much work is done in ploughing the length of the field?
Give reasons for the following. No work is done if a man is pushing against a wall.
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.
A max X goes to the top of a building by a vertical spiral staircase. Another man Y of the same mass goes to the top of the same building by a slanting ladder. Which of the two does more work against gravity and why?
Why is the work done on an object moving with uniform circular motion zero?
Fill in the boxe to show the corresponding energy transformation.

A body of mass 0.5 kg travels on straight line path with velocity v = (3x2 + 4) m/s. The net work done by the force during its displacement from x = 0 tox = 2 m is ______.
