A block of mass 2.0 kg is pushed down an inclined plane of inclination $37^{\circ}$ with a force of 20 N acting parallel to the incline. It is found that the block moves on the incline with an acceleration of $10 m / s^{2}$ . If the block started from rest, find the work done $(W_{F})$ by the applied force, work done $(W_{g})$ by the weight of the block and work done $(W_{f})$ by the frictional force acting on the block in the 1st second. Take $g = 10 m / s^{2}$ .

$W_{F} = 100 J; W_{g} = 60 J; W_{f} = 60 J$

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$W_{F} = 100 J; W_{g} = 60 J; W_{f} = - 60 J$

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$W_{F} = 100 J; W_{g} = 100 J; W_{f} = 0 J$

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$W_{F} = - 100 J; W_{g} = 100 J; W_{f} = 0 J$

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Solution

The correct option is B
$W_{F} = 100 J; W_{g} = 60 J; W_{f} = - 60 J$

So question is asking for work done in the first second
$a = 10 m / s^{2}$
$u = 0 m / s$
$t = 1 s e c$
So, in 1 sec displacement of block will be $s = u t + \frac{1}{2} a t^{2}$ $s = 5 m$
$W_{F}$ = Work done by force $= F \times s \times c o s θ$
$= 20 \times 5 \times c o s 0^{\circ}$
$= 100 J$
$W_{g}$ = Work done by weight $= m g \times s \times c o s θ$
$= 2 \times g \times 5 \times c o s 53^{\circ}$
$= 60 J$
$W_{f}$ = Work done by friction $= f r \times s \times c o s θ$
Oh, so we need NLM to find friction from free body drawing of object.
$F + m g c o s 53^{\circ} - f_{r} = m a$
$20 + 12 - f r = 2 \times 10$
$\Rightarrow f_{r} = 12 N$
$\Rightarrow W_{f} = 12 \times 5 \times c o s 180^{\circ} = - 60 J$

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