If a block moving up at θ = 30° with a velocity 5 m/s, stops after 0.5 sec, then what is μ

The correct answer is 0.6.

When a block moves up an inclined plane, two forces act down the incline simultaneously — the gravitational component along the slope and the kinetic friction force (since friction always opposes motion, it acts downward when the block moves up).

Step 1: Find the deceleration using kinematics.

Given:

Initial velocity: u = 5 m/s

Final velocity: v = 0 m/s (block stops)

Time: t = 0.5 s

Using the first equation of motion:

v=u+at

0=5+a\left(0.5\right)

a=-\frac{5}{0.5}=-10\text{m/s²}

So the magnitude of deceleration is 10 m/s².

Step 2: Apply Newton's Second Law along the incline (block moving up).

Forces acting down the incline:

① Gravitational component along slope = mg\sin \theta

② Friction force = \mu mg\cos \theta

Net retarding force = ma, so:

mg(\sin \theta +\mu \cos \theta )=m\times 10

Dividing both sides by m:

g(\sin \theta +\mu \cos \theta )=10

Step 3: Substitute values.

Using g = 10 m/s², θ = 30°, sin 30° = 0.5, cos 30° = \frac{\sqrt{3}}{2}:

10(0.5+\mu \times \frac{\sqrt{3}}{2})=10

0.5+\mu \times \frac{\sqrt{3}}{2}=1

\mu \times \frac{\sqrt{3}}{2}=0.5

\mu =\frac{0.5}{\frac{\sqrt{3}}{2}}=\frac{0.5\times 2}{\sqrt{3}}=\frac{1}{\sqrt{3}}

Step 4: Compute the numerical value.

\mu =\frac{1}{\sqrt{3}}=\frac{1}{1.732}\approx 0.577\approx 0.6

Therefore, the coefficient of friction μ ≈ 0.6.