A block slides down a curved frictionless surface of height $12 m$ and then moves up a rough inclined plane of angle of inclination $45^{o}$ . If the coefficient of kinetic friction between the block and the inclined plane is $0.2$ , the maximum height reached by the block is:

12 m

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10 m

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8 m

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6 m

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Solution

The correct option is C $10 m$
Here some potential energy is dissipated as work done against friction.
By work energy theorem:
$m g H = m g h + W_{f} = m g h + μ m g c o s θ (\frac{h}{s i n θ}) \Rightarrow h = \frac{m g H}{m g + μ m g c o s θ (\frac{1}{s i n θ})} = \frac{12}{1.2} = 10 m$

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A block slides down a curved frictionless surface of height 12m and then moves up a rough inclined plane of angle of inclination 45o. If the coefficient of kinetic friction between the block and the inclined plane is 0.2, the maximum height reached by the block is:

The correct option is C 10mHere some potential energy is dissipated as work done against friction. By work energy theorem: mgH=mgh+Wf=mgh+μmgcosθ(hsinθ)⇒h=mgHmg+μmgcosθ(1sinθ)=121.2=10m

A block slides down a curved frictionless surface of height $12 m$ and then moves up a rough inclined plane of angle of inclination $45^{o}$ . If the coefficient of kinetic friction between the block and the inclined plane is $0.2$ , the maximum height reached by the block is:

The correct option is C $10 m$
Here some potential energy is dissipated as work done against friction.
By work energy theorem:
$m g H = m g h + W_{f} = m g h + μ m g c o s θ (\frac{h}{s i n θ}) \Rightarrow h = \frac{m g H}{m g + μ m g c o s θ (\frac{1}{s i n θ})} = \frac{12}{1.2} = 10 m$