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300 J of work is done in sliding a 2 kg block up an inclined plane of height 10 m. Taking $g = 10$ m/s$^2$, work done against friction is
A
200 J
B
100 J
C
Zero
D
1000 J
Detailed Solution
By the work-energy theorem, $W_{net} = \Delta KE$. The block is slid up slowly, so $\Delta KE = 0$.
$\Rightarrow 300 - W_{gravity} - W_{friction} = 0$
Work done against gravity $= mgh = (2)(10)(10) = 200$ J
$\Rightarrow W_{friction} = 300 - mgh$
$= 300 - 200$
$= 100$ J
$\Rightarrow 300 - W_{gravity} - W_{friction} = 0$
Work done against gravity $= mgh = (2)(10)(10) = 200$ J
$\Rightarrow W_{friction} = 300 - mgh$
$= 300 - 200$
$= 100$ J