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Physics Question 44 – JEE-MAIN 2026

The velocity at which 6 kg mass (shown in figure) strikes the ground when it is released from a height of 6 m above the ground is _______ m/s. Assume pulley is massless and string is light and inextensible. (Take g=10 m/s2)

The problem can be solved efficiently by applying the principle of conservation of mechanical energy, as the system is conservative (only gravity and tension, with tension doing no net work on the system).

Step 1: Identify System and Apply Conservation of Energy✦ Active

The system consists of two masses (m1=6 kg and m2=2 kg) connected by a string over a massless, frictionless pulley. Since the string is inextensible and the pulley is massless, the tension forces do no net work on the system. Therefore, mechanical energy is conserved. The initial state has m1 at height h=6 m and m2 at ground level, both at rest. The final state has m1 at ground level and m2 at height h, both moving with velocity v.

Einitial=Efinal m1gh=m2gh+12(m1+m2)v2
Step 2: Solve for Final Velocity○ Expand

Rearrange the energy conservation equation to solve for the final velocity v.

(m1m2)gh=12(m1+m2)v2 v2=2(m1m2)ghm1+m2
💡 Teacher's Secret Hint

Ensure correct algebraic manipulation to isolate v2.

Step 3: Substitute Values and Calculate○ Expand

Substitute the given values: m1=6 kg, m2=2 kg, h=6 m, and g=10 m/s2 into the equation for v.

v=2(62)×10×66+2 v=2×4×10×68 v=4808=60 v7.7459 m/s

Rounding to two decimal places, v7.75 m/s. The closest option is 7.74 m/s.

💡 Teacher's Secret Hint

Pay attention to rounding instructions if specified, otherwise round to a reasonable number of significant figures or match the options.

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