Advertisements
Advertisements
प्रश्न
A body, when acted upon by a force of 10 kgf, gets displaced by 0.5 m. Calculate the work done by the force, when the displacement is (i) in the direction of force, (ii) at an angle of 60° with the force, and (iii) normal to the force. (g = 10 N kg-1)
Advertisements
उत्तर
Force acting on the body = 10 kgf = 10 × 10 N = 100 N
Displacement, S = 0.5 m
Work done = force x displacement in the direction of force
(i) W = F × S
W = 100 × 0.5= 50 J
(ii) Work = force x displacement in the direction of force
W = F × S cosθ
W = 100 × 0.5 cos60o
W = 100 × 0.5 × 0.5(cos 60o = 0.5)
W = 25 J
(iii) Normal to the force:
Work = force x displacement in the direction of force
W = F × S cosθ
W = 100 × 0.5 cos90o
W = 100 × 0.5 × 0 = 0 J(cos90o = 0)
संबंधित प्रश्न
State and define the S.I. unit of work.
A body of mass m is moving with a uniform velocity u. A force is applied on the body due to which its velocity changes from u to v. How much work is being done by the force?
What is a centripetal force?
Name the force needed for circular motion. What is the direction of the force?
An engine does 54,000 J of work by exerting a force of 6000 N on it. What is the displacement in the direction of the force?
State the condition when the work done by a force is negative. Explain with the help of examples.
1 kWh = ______ J.
In the following case write yes, if the work is being done and no if no work is being done.
A man trying to push a wall.
State the condition when the work done by the force is negative.
State the practical unit mainly used in mechanical engineering and write down its equivalent in watt.
