IIT JEE Entrance Exam − Profile

Step 1 Find Force F as a function of time t.

Force vs. time graph is linear:  F = C₁ t + C₂

To evaluate constants C₁ t + C₂ :

t = 0, F = 4 gives C₂ = 4, and

t = 3, F = 0 gives C₁ = −4/3.

So, F = −(4/3) t + 4.

Step 2 Find velocity v at time t = 4.5 s.

From Newton's 2nd law: F = m dv/dt

Integrating above equation gives

v = (1/m) ∫ F dt = (1/m) [−(4/3) t²/2 + 4 t]

Note v = 0 at t = 0.  Substituting m = 2 kg and t = 4.5 s,

v = ½ [−(2/3) (4.5)² + 4 (4.5)] = 2.25 m/s.

Step 3 Find kinetic energy KE at time t = 4.5 s.

KE = ½ m v² = ½ (2) (2.25)² = 5.0625 J.

Q = m C_p (T_final − T_initial) + m₁ λ

Heat added Q = 420 J

Heat capacity C_p = 2.1 J g^-1 °C^-1

Final temperature T_final = 0 °C

Initial temperature T_initial = −5 °C

Mass of ice melted m₁ = 1 g

Latent heat λ = 336 J g^-1

Substituting values gives

420 = m (2.1) (5) + (1) (336)

So, m = (420 − 336)/(2.1 × 5) = 8