Question

an object of 100 gm - weight is dropped from height of 2 metre from the ground calculate its velocity and kinetic energy when it reaches the ground [ g =10 m s -2]​

Answer 1

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Answer:

Kinetic Energy = 2J; Velocity = 6.325 $ms^{-1}$

Explanation:

Mass of the object = 100 grams = 0.1 kilograms

Height from the ground = 2 meter

Acceleration due to gravity = 10 $ms^{-2}$

$E_{Potential}$ = Mass × Gravitational acceleration  × Height

                                     = 0.1 × 10 × 2

                                     = 2J

Due to the Law of Conservation of Energy,

$E_{Potential}$ + $E_{Kinetic}$ = Constant

∵ The body is released from a height of 2 meters, it means that the body is at its maximum height, i.e. $E_{Potential}$ is maximum and $E_{Kinetic}$ is 0.

∴ $E_{Potential}$ + $E_{Kinetic}$ = Constant

  2J + 0J = 2J

∴ Constant = 2J

When the body reaches the ground, just before impact, $E_{Kinetic}$ will be maximum and $E_{Potential}$ will be 0

∴ $E_{Potential}$ + $E_{Kinetic}$ = Constant

   0J + $E_{Kinetic}$ = 2J

∴ $E_{Kinetic}$ = 2J

$E_{Kinetic}$ = $\frac{1}{2}$ × m × $v^{2}$

2J = $\frac{1}{2}$ × 0.1 × $v^{2}$

∴ $v^{2}$ = $\frac{2 * 2}{0.1}$

                          = $\frac{4}{0.1}$

                          = 4 ÷ 0.1

                          = 4 ÷ $\frac{1}{10}$

  $v^{2}$ = 4 × $\frac{10}{1}$ = 4 × 10 = 40

∴ $v$ = $\sqrt{40}$ = 6.325 $ms^{-1}$