Question

An object of 2 kg was placed at a height of 2 meter from the ground. A person liftsi up and keeps it at a height of 6 meter from the initial place. Calculate the change in potential energy of the object (g = 10 m/s²)​

Answer 1

$\text{Case I: Initial state}$

$\text{Height from ground}(h_1) = 2 \; m$

$\text{Mass of object}(m) = 2 \; kg$

$\text{Acceleration due to gravity}(g) = 10 \; m/s^{2}$

$\text{We know that Potential Energy}(U) = mgh$

$\implies U_1 = 2\times 10 \times 2$

$\implies U_1 = 40 \; J$

$\text{Case 2: Final state}$

$\text{Height from ground}(h_2) = (6 + 2) = 8 \; m$

$\implies U_2 = 2 \times 10 \times 8$

$\implies U_2 = 160 \; J$

$\implies \text{Change in Potential Energy}(\triangle{U}) = U_2 - U_1$

$\implies \triangle{U} = 160 - 40$

$\implies \boxed{\triangle{U} = 120 \; J}$