First slide

Mechanical energy conservation

Question

The system shown in Fig. is released from rest with mass 2 kg in contact with the ground. Pulley and spring are massless, and friction is absent everywhere. The speed of 5 kg block when 2 kg block leaves the contact with the ground is (force constant of the spring k = 40 N $m^{- 1}$ and g = 10 m $s^{- 2}$ )

Moderate

A

$\sqrt{2} m s^{- 1}$
B

$2 \sqrt{2} m s^{- 1}$
C

$2 m s^{- 1}$
D

3 $\sqrt{2} m s^{- 1}$

Solution

Let x be the extension in the string when 2 kg block leaves the contact with ground. Then tension in the spring should be equal to weight of 2 kg block:
Kx = 2g or $x = \frac{2 g}{K} = \frac{2 \times 10}{40} = \frac{1}{2} m$
Now from conservation of mechanical energy,
$m g x = \frac{1}{2} K x^{2} + \frac{1}{2} m v^{2}$
$\sqrt{2 g x - \frac{K x^{2}}{m}} = \sqrt{2 \times 10 \times \frac{1}{2} - \frac{40}{4 \times 5}} = 2 \sqrt{2} m s^{- 1}$

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