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
When a body is placed on a table top, it exerts a force equal to its weight downwards on the table top but does not move or fall
1) Name the force exerted by the table top
2) What is the direction of the force?
State the condition when the work done by a force is positive. Explain with the help of examples.
Differentiate between work and power.
kWh is the unit of ______.
An ox can apply a maximum force of 1000 N. It is taking part in a cart race and is able to pull the cart at a constant speed of 30 ms-1 while making its best effort. Calculate the power developed by the ox.
Give the units of couple.
State the S.I. and C.G.S. units of energy. How are they related?
A coolie carrying a load on his head and moving on a frictionless horizontal platform does no work. Explain the reason why.
A man raises a box of mass 50 kg to a height of 2 m in 20 s, while another man raises the same box to the same height in 50 s. Compare:
- the work done, and
- the power developed by them. Take g = 10 N kg-1
