JEE Main 2026 Apr 6 Shift 2, Physics Q19: Work Energy Theorem And Its Applications

JEE Main2026Apr 6, Shift 2Physics

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/s $^{2}$ ).

A
7.74
B
7.20
C
6.55
D
4.50

Solution

Use energy conservation. As the 6 kg block falls through $h = 6$ m, the 2 kg block rises by the same amount. Net gravitational work goes into kinetic energy of both blocks (same speed).

$(m_{1} - m_{2}) g h = \frac{1}{2} (m_{1} + m_{2}) v^{2}$

$(6 - 2) (10) (6) = \frac{1}{2} (8) v^{2}$ $\Rightarrow$ $240 = 4 v^{2}$ $\Rightarrow$ $v^{2} = 60$ .

$v = 60 \approx 7.746$ m/s.

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JEE Main 2026 Apr 6 Shift 2, Physics Q19: Work Energy Theorem And Its Applications

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