Physics [3 Mark Question Answer]

(a) Potential energy
(b) Potential energy
(c) Kinetic energy
(d) Potential energy
(e) Kinetic energy

The energy possessed by a body by virtue of its position, shape or change of configuration is called potential energy.
Examples of potential energy:
(i) Water stored at a height in a reservoir.
(ii) A stretched spring.
(iii) A bent bow.
The energy possessed by a body by virtue of its motion is called kinetic energy.
Examples of kinetic energy:
(i) Air in motion has kinetic energy.
(ii) A swinging pendulum.
(iii) Moving hands of a clock.

The energy of a body is its capacity to do work.
The SI unit of work is 'joules' and the CGS unit is 'erg'.
According to the law of conservation of energy, energy can neither be created nor be destroyed but can be transformed from one form to another. In other words, energy can be transformed from one form to another but the total amount of all the energies remain the same.

Let the weight 500 N be hung from the 40cm mark and the weight 300N be hung from the 80 cm mark

(i) Then, total clockwise moment = 500 x (50 - 40) = 500 x 10 = 5000 Nm

Total anticlockwise moment = 300 x (80 - 50) = 300 x 30 = 9000 Nm

(ii) Difference in clockwise and anticlockwise moment = 9000 - 5000 = 4000Nm

(iii) Let the 100 N weight be hung at a distance 'd' from the point 'o' to its left

Then,total clockwise moment = 5000 + 100 d

In balanced condition, sum of clockwise moments = sum of anticlockwise moments

5000 + 100 d = 9000

or 100 d = 4000

or d = 40cm

the weight 100 N should be a hung from the 40cm mark so as to balance the scale.

The body will move in the direction of net torque

Net torque = sum of clockwise moments - sum of anticlockwise moments

In the given figure, force 40N is acting clockwise and forces 20N and 60N are acting anticlockwise

Net torque = (40 x 3) - [(20 x 2) + (60 x 1)}

= 120 - 100

= 20 Nm

Hence the body will move in clockwise direction about 'O

In the given figure, forces 10 N and 100 N act clockwise and the forces 15 N and 4 N act anticlockwise

MOment of force = sum of clockwise moments - sum of anti clockwise moments

Moments of force = (10 x 5 + 100 x 0 + 4 x 6) - ( 15 x 4) = 74 - 60 = 14 Nm

or 14 Nm in clockwise direction.

When the centre of gravity is nearer to the base of a body, the body is in stable equilibrium.
Conditions for stable equilibrium:
(a) The body should have a broad base.
(b) Centre of gravity of the body should be as low as possible.
(c) Vertical line drawn from the centre of gravity should fall within the base of the support.

Equilibrium in any case requires the ? forces acting on an object = 0, i.e. that there is
Fnet = 0.
Static equilibrium is the situation where the object upon which the forces act is no moving.
The object is "static" hence the state of equilibrium gets its name.
Dynamic equilibrium is the situation where an object is in constant velocity motion.
{This object can't experience an acceleration which means Fnet >0}

The turning effect produced by a force on a rigid body about a point, pivot or fulcrum is called the moment of force or torque. It is measured by the product of force and the perpendicular distance of the pivot from the line of action of force.
Examples of turning effect of force:
(i) Turning a steering wheel
(ii) Tightening a cap

Centripetal force: Whenever a body is moving in a circular path with a uniform speed, its velocity is continuously changing due to change in its direction. The body thus possesses acceleration and this acceleration is called centripetal acceleration. The force which produces this acceleration is called centripetal force. It acts along the radius towards the centre of the circular path.
It is not the same as the centrifugal force.

A body is said to be in equilibrium under the action of a number of forces, if the forces are not able to produce any change in the state of rest or of uniform motion or uniform rotation.
Conditions for stable equilibrium:
(a) The body should have a broad base.
(b) Centre of gravity of the body should be as low as possible.
(c) Vertical line drawn from the centre of gravity should fall within the base of the support.

Let the mass of 50 g be situated at distance 'd' from the mid-point i.e. at 50cm,

80 x 30 = (40 x 10) + (50 x d)

2400 = 400 + 50d

or 50d = 2000

or d = 40cm to the right of the mid-point.

(i) At the height H because the height is maximum.
(ii) At the ground level because the velocity is maximum.
(iii) At half distance of the total path i.e. at height H/2, the P.E. is equal to the K.E.

(i) Potential energy
(ii) Heat energy
(iii) Potential energy

Potential energy of a body depends upon:
(i) Mass of the body
(ii) Acceleration due to gravity
(iii) Height of the body

(i) The conditions for static equilibrium are:
(a) The sum of the (vector) forces must equal zero, i.e. ? F = 0
(b) The sum of the torques must equal zero; i.e. ? ? = 0.
(ii) The conditions for dynamic equilibrium are:
(a) The body should have a broad base.
(b) Centre of gravity of the body should be as low as possible.
(c) Vertical line drawn from the centre of gravity should fall within the base of the support.
Examples:
Static equilibrium: a box at rest on a floor; there is a gravitational force pulling the object to the earth, but there is also an equal and opposite force applied by the floor to the box (pushing up).
Dynamic equilibrium: A rock travelling across the cosmos, far enough away from any other object (so as not to be affected by gravity - in other words, in zero gravity conditions); the rock continues to travel in a straight line at uniform velocity either for eternity, or until acted upon by an external unbalanced force.

Conditions of equilibrium for a rigid body:
1. The body should have a broad base.
2. Center of gravity of the body should be as low as possible.
3. Vertical line drawn from the center of gravity should fall within the base of
support.
4. Vector sum of forces acting on the body should be zero.
5. Algebraic sum of moments acting on the body should be zero.