Question

12. A body of mass m is launched up on a rough inclined plane making an angle 45° with horizontal . If the
time of ascent is half of the time of descent, the frictional coefficient between plane and body is

​

Answer 1

Answer:

The coefficient of friction between the plane and the body is 3/5

Explanation:

Let the friction coefficient be between the plane and the body be μ

Let the initial velocity with which the body is launched is u

Now, as shown in the attached diagram, the weight of the body can be resolved into two components.

Balancing the forces, perpendicular to the plane we get

The normal reaction $N={mg}\cos45^\circ$

The force of friction F = μN = ${mg\mu}\cos45^\circ$

Let the acceleration when the body is going up be a₁ and the acceleration when the body is going down be a₂

When the body is going up, if it covers a total distance s and if the time of ascent is t₁ then

By the first equation of motion

$0=u-a_1t_1$ (since acceleration will be in downwards direction, hence negative)

⇒ $a_1t_1=u$                                         .......... (1)

Also, when the body is going up, using the second equation of motion

$s=ut_1-\frac{1}{2}a_1t_1^2$                    ........ (2)

When the body is coming down, then it will again cover a total distance s and if the time of descent is t₂ then again using second equation of motion

$s=0\times t_1+\frac{1}{2}a_2t_2^2$

or, $s=\frac{1}{2}a_2t_2^2$

But $t_1=\frac{1}{2}t_2$

or, $t_2=2t_1$

∴ $s=\frac{1}{2}a_2(2t_1)^2$

or, $s=2a_2t_1^2$                               ........... (3)

From (1), (2) and (3)

$2a_2t_1^2=a_1t_1\times t_1-\frac{1}{2}a_1t_1^2$

or, $2a_2t_1^2=a_1t_1^2-\frac{1}{2}a_1t_1^2$

or, $2a_2t_1^2=\frac{1}{2}a_1t_1^2$

or, $a_1=4a_2$

Now, from Newton's second law of motion

When the body is going up

$F+{mg}\sin45^\circ=ma_1$

or, $\mu{mg}\cos45^\circ+{mg}\sin45^\circ=ma_1$

or, $a_1=\mu{g}\frac{1}{\sqrt{2} } +{g}\frac{1}{\sqrt{2} }$

or, $a_1=(\mu+1)\frac{g}{\sqrt{2}}$

Similarly, when the body is going down

${mg}\sin45^\circ-F=ma_2$

or, ${mg}\sin45^\circ-\mu{mg}\cos45^\circ=ma_2$

or, $a_2=(1-\mu)\frac{g}{\sqrt{2}}$

∵ $a_1=4a_2$

∴ $(\mu+1)\frac{g}{\sqrt{2}}=4\times(1-\mu)\frac{g}{\sqrt{2}}$

or, $5\mu=3$

or, $\mu = \frac{3}{5}$

Answer 2

Answer:

Explanation:

td=xta

2Lg(sinθ−μcosθ)=r2Lg(sinθ+μcosθ)

∴  μ=(x2−1x2+1)tanθ

= (22−122+1)=35