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Engineering Mechanics

Conservative Forces and Potential Energy

A block of ma...

Question

A block of mass of 5 kg is released from rest from a position A on an inclined plane as shown on the figure. If the coefficient of the friction between the block and the plane is 0.2, the velocity at which it will touch the ground at B is $(g = 10 m / s^{2})$

10 m/s

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15 m/s

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20 m/s

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24 m/s

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Solution

The correct option is C 20 m/s
$\frac{1}{2} m (V_{B}^{2} - V_{A}^{2}) = m g h - μ m g c o s θ . l$

$\frac{1}{2} (V_{B}^{2}) = g h - μ g x$

$V_{B} = \sqrt{2 (h - μ x) g} = 20 m / s e c$

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A block of mass of 5 kg is released from rest from a position A on an inclined plane as shown on the figure. If the coefficient of the friction between the block and the plane is 0.2, the velocity at which it will touch the ground at B is (g=10m/s2)

The correct option is C 20 m/s12m(V2B−V2A)=mgh−μmgcosθ.l 12(V2B)=gh−μgx VB=√2(h−μx)g=20m/sec

A block of mass of 5 kg is released from rest from a position A on an inclined plane as shown on the figure. If the coefficient of the friction between the block and the plane is 0.2, the velocity at which it will touch the ground at B is $(g = 10 m / s^{2})$

The correct option is C 20 m/s
$\frac{1}{2} m (V_{B}^{2} - V_{A}^{2}) = m g h - μ m g c o s θ . l$

$\frac{1}{2} (V_{B}^{2}) = g h - μ g x$

$V_{B} = \sqrt{2 (h - μ x) g} = 20 m / s e c$