Question

If a body of mass 2kg is projected vertically upwards with a velocity of 20 m/s then find potential energy after 2s?​

Answer 1

Given ,

Mass of body ,m=2 kg

Initial velocity u=20m/s

Height reached at t = 2secs h = ut - 1/2gt^2

⟹ h = 20 × 2 − 1/2 × 10 × 2^2

⟹ h = 20 m

Potential energy , P=mgh=2×10×20=400 J

Answer 2

Question:-

• If a body of mass 2kg is projected vertically upwards with a velocity of 20 m/s then find potential energy after 2s?

Solution:-

• Given:

↝Mass (m)→2kg

↝Velocity (u)→20m/s

↝Time (t)→2s

Let us assume the height be (h)

• To find:

☞Potential energy after 2 second

• Here:

By applying formula of height (h):

$➝h = ut - \frac{1}{2} g {t}^{2}$

Let gravitational constant (g) be = -10m/s²

• Minus '-' sign is used because ball is thrown vertically upwards i.e. against the gravity.

$➝h =( 20 \times 2) - \frac{1}{2} \times 10 \times {2}^{2}$

$➝h = 20m$

• Here we got height=20m

Now:-

Potential Energy of an object ➝mgh

• m=maas→2kg
• g=gravitational constant→10m/s²
• h=height→20m

Therefore:-

Potential energy after 2s=(2×10×20)J

Potential energy after 2s=400J

• Hence potential energy after 2s of this object will be 400J

Hope it helps you ✔️✔️

@Fabled