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A block has an initial kinetic energy of 128 J. It slides up from point A at the bottom of the inclined plane with uniform deceleration. When it passes point B, its kinetic energy is reduced by 80J, and its mechanical energy is reduced by 35 J. Work done against friction when the block moves from A to the highest point C on the inclined plane is given by $14 α$ . Then find the value of $α$ .

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Detailed Solution

At A Total energy=128 J
At B K=48 J, u=45 J $W_{f} = 35 J$
$U = m g . X \sin θ = 45 W_{f} = f x = 35 \frac{m g x \sin θ}{f x} = \frac{45}{35} f = \frac{7}{9} m g \sin θ$

Now body stops at C.
Let y=total displacement on the plane
So, $m g \sin θ + f y = 128$
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