Question

⚡⚡A mass of 10 kg is dropped from a height of 50 cm. Find its -

a) potential energy

b) kinetic energy just on touching the ground

c) velocity with which it hits the ground (g = 10m/s²)​

Answer 1

Question -

A mass of 10 kg is dropped from a height of 50 cm. Find its -

a) potential energy

b) kinetic energy just on touching the ground

c) velocity with which it hits the ground (g = 10m/s²)

Solution :

Mass = m = 10kg

Initial height = h = 50cm

Converting into m = 0.5 m

g = 10m/s² (given)

a) Potential Energy of Mass just before dropping =

mgh = 10×10×0.5 = 50 J

b) Kinetic Energy of mass just on touching the ground = Potential Energy of mass just before dropping = 50J

c) If final Velocity with the mass hits the ground be v, then

$e_{k} = \frac{1}{2} m {v}^{2} \\ \\ v = \sqrt{ \frac{2 e_{k} }{m} } \\ \\ = \sqrt{ \frac{2 \times 50}{10} } = 3.16m {s}^{ - 2}$

Answer 2

Mass of The Body=10 kg

Height from its Dropped=50cm

(1)Potenial Energy=Mass×g×Height.

Gravitational Accerlation=10 m/s

Height=50 cm =0.5 m

Mass=10 kg

Potenial Energy=10×10×0.5

=100×0.5

=50 J

$\boxed{\mathbf{PE=50\:J}}$

(2)KE=1/2mv²

At Just Before Touching the Ground 50J energy is also Acting.

Then KE=50J

v=Final Velocity

1/2×10×v²=50J

5×v²=50

v²=50/5

v²=10

v=√10 m/s

$\boxed{\mathbf{KE=50\:J}}$

(3)√10 m/s is velocity just on Touching.

So Final Velocity=√10 m/s

$\boxed{\mathbf{Velocity=\sqrt{10}m/s}}$