Work Energy Power

JEE Main level, single correct, medium

35 of 183

Question 35 of 183

The velocity at which 66 kg mass (shown in figure) strikes the ground when it is released from a height of 66 m above the ground is __________ m/s. Assume pulley is massless and string is light and inextensible. (Take g =10= 10 m/s2^2)

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Answer: (A) 7.747.74

Let the acceleration of the system be aa and the tension in the string be TT.

The equations of motion for the two masses are:
6g−T=6a6g - T = 6a
T−2g=2aT - 2g = 2a

Adding the two equations, we get:
4g=8a4g = 8a
⇒a=4g8=g2\Rightarrow a = \dfrac{4g}{8} = \dfrac{g}{2}

Substituting g=10 m/s2g = 10 \text{ m/s}^2, the acceleration is:
a=102=5 m/s2a = \dfrac{10}{2} = 5 \text{ m/s}^2

The 6 kg6 \text{ kg} mass is released from rest (u=0u = 0) and travels a downward distance s=6 ms = 6 \text{ m}.

Using the kinematic equation v2=u2+2asv^2 = u^2 + 2as:
v2=0+2×5×6v^2 = 0 + 2 \times 5 \times 6
v2=60v^2 = 60
v=60≈7.74 m/sv = \sqrt{60} \approx 7.74 \text{ m/s}

Answer: 7.747.74

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