$300 J$ of work is done in sliding a $2 k g$ block upon an inclined plane of height $10 m$ .Work done against friction is :-
( take $g = 10 m / s^{2}$ )

z e r o

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100 J

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200 J

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300 J

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Solution

The correct option is A $100 J$
Let work done against the friction is $W_{f}$ .

Work done by gravity $W_{g} = m g h = 2 \times 10 \times 10 = 200 J$

Using Work-Energy theorem, Total work done = Change in kinetic energy

Here $Δ K E = 0$ ,

$\Rightarrow 300 - W_{g} - W_{f} = 0$ $\Rightarrow W_{f} = 100$ $J$

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300 J of work is done in sliding a 2 kg block upon an inclined plane of height 10m.Work done against friction is :-( take g=10m/s2)

The correct option is A 100 JLet work done against the friction is Wf.Work done by gravity Wg=mgh=2×10×10=200JUsing Work-Energy theorem, Total work done = Change in kinetic energyHere ΔKE=0, ⇒300−Wg−Wf=0 ⇒Wf=100 J

$300 J$ of work is done in sliding a $2 k g$ block upon an inclined plane of height $10 m$ .Work done against friction is :-
( take $g = 10 m / s^{2}$ )