A block starts moving up an inclined plane of inclination 30° with an initial velocity of v0. It comes back to its initial position with velocity v₀/2. The value of the coefficient of kinetic friction between the block and the inclined plane is close to I/1000. The nearest integer to I is____. Numerical answer type. Work it out, then open the answer.

Answer: 346. Up a distance d: 1/2mv₀² = mgd(sin 30° + μcos 30°). Down: 1/2m(v₀²)/(4) = mgd(sin 30° - μcos 30°). Divide: 4 = (sin 30° + μcos 30°)/(sin 30° - μcos 30°) ⇒ μ = 3/5tan 30° μ = (0.6)/(√(3)) = 0.346 = (346)/(1000), so I = 346.

A block starts moving up an inclined plane of inclination 30° with an… | JEE Main 2020 (3 Sep, Shift 2)

Q29JEE Main 2020 · 3 SEP · SHIFT 2Difficult

Conservation of Energy

A block starts moving up an inclined plane of inclination 30° with an initial velocity of v0. It comes back to its initial position with velocity $\frac{v_0}{2}$ . The value of the coefficient of kinetic friction between the block and the inclined plane is close to $\frac{I}{1000}$ . The nearest integer to I is____.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 346

Up a distance $d$ : $\frac{1}{2}mv_0^2 = mgd(\sin 30^\circ + \mu\cos 30^\circ)$ .
Down: $\frac{1}{2}m\frac{v_0^2}{4} = mgd(\sin 30^\circ - \mu\cos 30^\circ)$ .
Divide: $4 = \frac{\sin 30^\circ + \mu\cos 30^\circ}{\sin 30^\circ - \mu\cos 30^\circ} \Rightarrow \mu = \frac{3}{5}\tan 30^\circ$
$\mu = \frac{0.6}{\sqrt{3}} = 0.346 = \frac{346}{1000}$ , so $I = 346$ .

Report a mistake

Conservation of Energy question from JEE Main 2020 (3 Sep, Shift 2)

More questions on Conservation of Energy