A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of 60 by a force of 10 N parallel to the inclined surface as shown in the figure. When the block is pushed up by 10 m along the inclined surface, the work done against friction

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Solution

Q.
A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of 60 by a force of 10 N parallel to the inclined surface as shown in the figure. When the block is pushed up by 10 m along the inclined surface, the work done against frictional force is $[g = 10 m / s^{2}]$ [2024]

1 $5 \sqrt{3}$ J

2 5 J

3 $5 \times 10^{3}$ J

4 10 J

Ans.
(2)

$f_{k} = μ_{k} N = μ_{k} m g \cos 60 °$

$f_{k} = 0.1 \times 10 \times \frac{1}{2} = 0.5 N$

$W_{friction} = - f_{k} \cdot S$

$= - (0.5) \times 10$

$W_{friction} = - 5 J$

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