JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Work, Power & Energy

A ball of mass 100 g is projected with velocity

20 m / s

6 0^{\circ}

with horizontal. The decrease in kinetic energy of the ball during the motion from point of projection to highest point is :

(A)20 J
(B)15 J✓ Correct
(C)zero
(D)5 J

Answer: (B)

Solution

Initial kinetic energy at launch (v = 20 m/s, m = 0.1 kg):

K_{i} = \frac{1}{2} m v^{2} = \frac{1}{2} (0.1) (400) = 20 J

At the highest point of the trajectory, only the horizontal component of velocity survives:

v_{x} = v cos 60° = 20 \times \frac{1}{2} = 10 m/s

K_{f} = \frac{1}{2} m v_{x}^{2} = \frac{1}{2} (0.1) (100) = 5 J

Decrease in kinetic energy:

Δ K = K_{i} - K_{f} = 20 - 5 = 15 J

JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Single correct option

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