Question 557

The upper half of an inclined plane of inclination θ is perfectly smooth while the lower half rough. A block starting from rest at the top of the plane will again come to rest at the bottom if the coefficient of friction between the block and the lower half of the plane is given by

μ = 2 tanθ

μ = tanθ

μ = 2/(tan θ)

μ = 1/ tan θ

Solution

A.

μ = 2 tanθ

suppose the length of plane is L. When block descends the plane, rise in kinetic energy = fall in potential energy = mg L sinθ i

Work done by friction = μ x (reaction) x distance

0 + μ (mg cos θ) x (L/2)

= μ (mg cos θ) x (L/2)

Now, work done = change in KE

mgL sin θ = μ (mg cos θ) x (L/2)

⇒ tan θ = μ/2

μ = 2 tan θ

Question 558

Two masses 10 kg and 20 kg respectively are connected by a massless springs as shown in the figure. A force of 200 N acts on the 20 kg mass. At the instant shown is a figure the 10 kg mass has an acceleration of 12 m/s2. The value of the acceleration of 20 kg mass is

4 m/s

^{2}10 m/s

^{2}20 m/s

^{2}30 m/s

^{2}

Solution

A.

4 m/s^{2}

The equation of motion of m_{1} = 10 kg mass is

F_{1} = m_{1}a_{1} = 10 x 12 = 120 N

Force on 10 kg mass is 120 N to the right. As action and reaction are equal and opposite, the reaction force F- on 20 kg mass F = 120 N to the left.

therefore, equation of motion of mass m2 = 20 kg is

200 - F = 20 a_{2}

200-120 = 20a_{2}

80 = 20a_{2}

a_{2} = 80 /20 = 4 m/s^{2}

Question 559

A cylinder rolls up an inclined plane, reaches some height and then rolls down (without slipping throuhgout these motions). The directions of the frictional force acting on the cylinder are

Up the incline, while ascending and down the incline while descending

Up the incline, while ascending as well as descending

down the incline, while ascending and up the incline while descending

Down the incline while ascending as well as descending.

Solution

B.

Up the incline, while ascending as well as descending

It is obvious that during ascending, a retarding i.e. anticlockwise moment is required. I should be remembered that torque due to friction has the same sense the angular acceleration.

Question 560

A long block A of mass M is at rest on a smooth horizontal surface. A small block B of mass M/2 is placed on A at one end and projected along A with same velocity v. The coefficient of friction between the block is μ. Then the acceleration of blocks A and B before reaching a common velocity will be respectively

$\frac{\mathrm{\mu g}}{2}(\mathrm{towards}\mathrm{right}),\frac{\mathrm{\mu g}}{2}(\mathrm{towards}\mathrm{left})$

$\mathrm{\mu g}(\mathrm{towards}\mathrm{right}),\mathrm{\mu g}(\mathrm{towards}\mathrm{left})$

$\frac{\mathrm{ug}}{2}(\mathrm{towards}\mathrm{right}),\mathrm{\mu g}(\mathrm{towards}\mathrm{left})$

$\mathrm{\mu g}(\mathrm{towards}\mathrm{right}),\frac{\mathrm{\mu g}}{2}(\mathrm{towards}\mathrm{left})$

Solution

B.

$\mathrm{\mu g}(\mathrm{towards}\mathrm{right}),\mathrm{\mu g}(\mathrm{towards}\mathrm{left})$

The force causing the motion of A is a frictional force between A and B,

So, acceleration of A

${\mathrm{\mu M}}_{\mathrm{B}}\mathrm{g}={\mathrm{M}}_{\mathrm{A}}{\mathrm{a}}_{\mathrm{A}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{a}}_{\mathrm{A}}=\mathrm{\mu}\left(\frac{{\mathrm{M}}_{\mathrm{B}}}{{\mathrm{M}}_{\mathrm{A}}}\right)\mathrm{g}=\frac{\mathrm{\mu g}}{2}(\mathrm{towards}\mathrm{right})$

Block B experiences frictions force towards the left.

MBaB = μMBg ⇒a_{B} = μg towards left

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