Olympiad Test: Work and Energy- 1 - Class 9 MCQ

perpendicular (at 90°) to the direction of force

(gravitational weight mg in this case), no

work is done.

W = F cosθ ×s

= F × cos 90° × s

= 0 (cos 90° = 0)

As the body comes down the height of the body decreases and hence the potential energy decreases. If the potential energy decreases the kinetic energy will increase.

As soon as the body touches the ground, all the initial potential energy of the particle gets converted into kinetic energy.

The potential energy (P.E) of a body at a certain height is 200 J. The kinetic energy possessed by it when it just touches the surface of the earth is the same as the Potential Energy (= P.E.). So the option C is correct.

W = m g h

= 50 × 10 × 200/100

= 1000 J

W = f cos 60°× d

= 15 × ½ × 20

= 150 Nm

= 150 J

= 0.01 J

P.E. = mgh

480 J = 12 kg × 10 ms^–2 × h

h = 480 J/120 kg ms^-2

= 4 m

The object is at the height of 4 m above the ground.

K.E. = ½ mv²

p= mv => K.E. = ½ m (p/m)² = p²/2m

∴ K.E₁/K.E₂ = (P₁²/2m₁)/(p₂²/2m₂) = m₂/m₁ = p₁²/p₂²

Here 'p₁' = p₂ (given)

∴ K.E₁/K.E₂ = 4×10^-3 kg/ 1×10^-3 kg = 4/1

The correct option is B becomes four times

KE =½ mv²

Given, v′=2v

KE′ =½ mv′² = 1/2m × 4 × v²

From the above two equations we get:

KE′= 4 × KE

Hence, kinetic energy is quadrupled.

The correct option is (C) Kinetic energy

E_T = KE + PE

at mean position : E_T = KE_max + 0

so E_T = KE_max

The mean position of the pendulum is its lowest point. Now we know that energy will be conserved because of the law of conservation of energy. This means that at the extreme position, the pendulum will have maximum potential energy. As the pendulum swings down, its potential energy will decrease and it's kinetic will increase. So if it is at the mean position all the potential energy will be converted into kinetic energy. (Assuming the SHM preforms motion under ideal condition and no PE is imparted on the body at mean position.

P.E. = mgh

= 100 kg × 10 m/s² × 50/100 m

= 500 J