Question

1. Calculate the work done when a force of 60 N
displaces a body through 10 m in the direction
of the force.
(Ans. 600 J)
2. When a force of 30 N acts on a body, the body
is displaced by 10 m in the direction inclined
at 60° to the force. Find the work done by the
force.
(Ans. 150 J)
3. A body of mass 5 kg is lifted through a height
of 20 m. Calculate the work done by the
applied force.
(Ans. 980 J)
4. If the energy of a ball falling from a height of
20 m is reduced by 60%, how high will it
rebound?
(Ans. 8 m)
5. A body of mass 2 kg moves with a velocity of
5 m/s. Find its kinetic energy. (Ans. 25 J)
6. The speed of a body of mass 4 kg increases
from 2 m/s to 4 m/s in a certain time interval.
Find the increase in the kinetic energy of the
body.
(Ans. 24 J)
7. Calculate the potential energy of a body of
mass 5 kg at a height of 2 m from the ground.
(Ans. 98 J)​

Answer 1

Answer:

1.f×d cos theta

befikra is a angle between the force and displacement but from the question it is clear that the force and displacement are in the same direction hill the angle between them will be zero that means one hence f into that is 60 into 10 will be equal to 600 Newton

Answer 2

Answer:

1. work done is 600J

2.work done is 150J

3. work done is 1960 J

4. rebound height is 8m

5.kinetic energy is 25J

6. increase in kinetic energy is 24J

7. potential energy is 98J

Explanation:

1. given, force = 60 N, and displacement = 10 m

work = force× displacement

work done = 60 × 10 = 600J

2. given, force = 30N, displacement = 10m , inclination angle, Θ = 60°

work done = force× displacement× cosΘ

             = 30× 10 × cos 60

             = 150J

3. given, mass = 5kg, height = 20 m

force = mass×gravity = 10× 9.8 = 98 N

work done = 98× 20 = 1960 J

4. given, height = 20m, reduced energy = 60%

let total energy be 100%

then, remaining energy = 100-60 = 40%

potential energy = mass× gravitational field× height = m×9.8×20 = 196 m

rebound height = potential energy

let rebound height be h'

then, mgh' = 196m× 40%

h' = $\frac{196(40)}{9.8(100)}$ = 8m

5. given, mass = 2 kg, velocity = 5m/s

kinetic energy = $\frac{mv^{2} }{2}$ = $\frac{(2)5^{2} }{2}$ = 25J

6. given, mass = 4 kg, speed increases from 2 m/s to 4 m/s

increase in kinetic energy = $\frac{m(v^{2}-u^{2}) }{2}$ = $\frac{4(4^{2}-2^{2}) }{2}$ = 24J

7. given, mass = 5 kg, height = 2m

potential energy = mass× gravitational field× height = 5× 9.8× 2 = 98J