First slide

Work done by Diff type of forces

Question

A block B of mass 2 kg is placed on the rough inclined surface of a wedge W. The wedge is kept on a smooth horizontal floor and held in position by an external agent. The block is in equilibrium on the wedge. If the wedge slowly moves to the left by 5m,

Difficult

A

Work done by force of friction on the block is $- 20 \sqrt{3} J$
B

Work done by normal force acting on the block is $+ 25 \sqrt{3} J$
C

Work done by force of friction on the block is $+ 25 \sqrt{3} J$
D

None of the above is correct.

Solution

$f = mgsinθ and N = mgcosθ$
$\begin{array}{l} ∴ f = 2 \times 10 \times \sin 30^{o} N = 10 N \\ N = 2 \times 10 \times \cos 30^{o} N = 10 \sqrt{3} N \end{array}$
$\begin{array}{l} ∴ W_{f} = - (fcosθ) S = - 10 \times \cos 30^{o} \times 5 J = - 25 \sqrt{3} J \\ W_{N} = Nsinθ \times S = 10 \sqrt{3} \times \sin 30^{o} \times 5 J = + 25 \sqrt{3} J \end{array}$

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