Question

A stone of mass 50 g is thrown vertically upward from the ground with the initial velocity of 20.0 ms⁻¹. Gravitational potential energy at ground level is considered zero. Apply the principle of conservation of energy to calculate the potential energy at the maximum height attained by stone. [g = 10 ms⁻¹]

Answer 1

At the maximum height, the stone's velocity becomes zero, so its kinetic energy is zero. By the law of conservation of energy, its initial kinetic energy at the ground is converted entirely into gravitational potential energy at the maximum height.

1. Mass conversion into S.I. unit:

\[ m = 50\ \text{g} = \frac{50}{1000}\ \text{kg} = 0.050\ \text{kg} \]

2. Initial Kinetic Energy (K_i) at the ground:

\[ K_{i} = \frac{1}{2}mv^{2} \]

\[ K_{i} = \frac{1}{2} \times 0.050 \times (20.0)^{2} \]

\[ K_{i} = \frac{1}{2} \times 0.050 \times 400 = 10\ \text{J} \]

3. Potential Energy (U) at maximum height:

According to conservation of energy:

\[ U = K_{i} = {10\ J} \]