First slide

Work Energy Theorem

Question

A 4 kg block is projected with velocity 7 m/s up a rough inclined surface having inclination 45^o with horizontal. The block stops after travelling a distance of $2 \sqrt{2}$ m along the plane. Then coefficient of kinetic friction between the block and the plane is

Moderate

A

0.750
B

0.225
C

0.125
D

0.425

Solution

Resultant force acting on the block, $F_{n e t} = m g \sin θ + μ (m g \cos θ)$
$∴$ Net work done on the block
$W_{net} = {\vec{F}}_{net} . \vec{S} = - mg (sinθ + μcosθ) . S .$
By work - energy theorem, $W_{net} = ΔK$
$∴ - m g (\sin θ + μ \cos θ) S = 0 - \frac{1}{2} m V^{2}$
$\Rightarrow μ = \frac{V^{2}}{2 g s \cos θ} - \tan θ = \frac{{(7)}^{2}}{2 \times 10 \times 2 \sqrt{2} \times \cos 45^{o}} - \tan 45^{o}$
$\Rightarrow μ = 0.225$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App