First slide

Friction on inclined plane of angle more than angle of repose

Question

A block of mass 5 kg slides down a rough inclined surface. The angle of inclination is 45⁰. The
coefficient of sliding friction is 0.20 . When the block slides 10 cm, the work done on the block by
force of friction is

Moderate

A

$- \frac{1}{\sqrt{2}} J$
B

$1 J$
C

$- \sqrt{2} J$
D

$- 1 J$

Solution

Work done by frictional force, $W_{f} = f s \cos 180 °$
$= (μ m g \cos θ) (- 1) (s)$
$= - 0.20 \times 5 \times 10 \times \frac{1}{\sqrt{2}} \times 0.1 (∵ θ = 45 °)$
$= - \frac{1}{\sqrt{2}} J$

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