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Standard XII

Physics

Law of Conservation of Mechanical Energy

A 4 kg block ...

Question

A $4 kg$ block is on a smooth horizontal table. The block is connected to a second block of mass $1 kg$ by a massless flexible taught cord that passes over a frictionless pulley. The $1 kg$ block is $1 m$ above the floor. The two blocks are released from rest. With what speed (in $m/s$ ) does the $1 kg$ block hit the ground?

3 m/s

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4 m/s

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2 m/s

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5 m/s

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Solution

The correct option is C $2 m/s$
By the law of mechanical energy conservation
Decrease in potential energy of $1 kg$
= increase in kinetic energy of both
$∴ 1 \times 10 \times 1 = \frac{1}{2} (4 + 1) v^{2}$
(speed will be same as they are connected by string )
$v = 2 m/s$

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The correct option is C 2 m/sBy the law of mechanical energy conservation Decrease in potential energy of 1 kg = increase in kinetic energy of both ∴1×10×1=12(4+1)v2 (speed will be same as they are connected by string ) v=2 m/s

The correct option is C $2 m/s$
By the law of mechanical energy conservation
Decrease in potential energy of $1 kg$
= increase in kinetic energy of both
$∴ 1 \times 10 \times 1 = \frac{1}{2} (4 + 1) v^{2}$
(speed will be same as they are connected by string )
$v = 2 m/s$