Question

A stone of mass m=0.5kg is projected up at an angle of 30° with the horizontal, with
an initial velocity of = 20ms. The work done by the force of gravity in the first two
second is : (g=10ms=2)
1) 50 J
2) 25 J
3) -25 J
4) zero
​

Answer 1

Given :

Mass of the stone = 0.5 kg

Angle of inclination = 30°

Initial velocity = 20 m/s

To Find :

The work done by the force of gravity in the first two seconds.

Solution :

❖ We know that work done is measured as the product of force and displacement.

• So first of all we have to find displacement of stone after 2 second.

Since gravitational force acts in the downward direction, we will use vertical component of velocity to find displacement of stone.

Displacement of stone after 2 seconds is given by

$\sf:\implies\:d=(u_y\times t)+\dfrac{1}{2}gt^2$

• $\sf{u_y=u\:sin\theta}$

$\sf:\implies\:d=(u\:sin\theta\times t)+\dfrac{1}{2}(-g)t^2$

$\sf:\implies\:d=(20\:sin30^{\circ}\times 2)-\dfrac{1}{2}(10\times 2^2)$

$\sf:\implies\:d=(40\times 0.5)-\dfrac{1}{2}(40)$

$\sf:\implies\:d=20-20$

$\bf:\implies\:d=0\:m$

Hence work done by force of gravity will be zero.

Answer 2

Given :

Mass of the stone = 0.5 kg

Angle of inclination = 30°

Initial velocity = 20 m/s

To Find :

The work done by the force of gravity in the first two seconds.

Solution :

❖ We know that work done is measured as the product of force and displacement.

So first of all we have to find displacement of stone after 2 second.

Since gravitational force acts in the downward direction, we will use vertical component of velocity to find displacement of stone.

Displacement of stone after 2 seconds is given by

: ⟹ d = (u^y × t) + 1/2 gt²

• u^y =u sinθ

: ⟹ d = (u sinθ × t) + 1/2 (-g)t²

: ⟹ d = (20 sin30∘ × 2) − 1/2( 10 × 2²)

: ⟹ d = (40 × 0.5) − 1/2(40)

: ⟹ d = 20 − 20

: ⟹ d = 0m

Hence, work done by force of gravity will be zero.