Question

A block of mass M is placed over incline of ${37}^{\circ }$ and coefficient of friction is 0.8 inside an elevator car, then

• A. Minimum acceleration of elevator so that block slides over plane is $gm/s^2$
• B. Direction of acceleration is upwards
• C. Direction of acceleration is downwards
• D. If acceleration of elevator is $\frac{g}{11}m/s^2$ upwards work done by friction on block will be positive

Answer 1

Answer: (A), (C), (D)

The correct options are
A Minimum acceleration of elevator so that block slides over plane is $gm/s^2$
C Direction of acceleration is downwards
D If acceleration of elevator is $\frac{g}{11}m/s^2$ upwards work done by friction on block will be positive

Case1:

The acceleration(a) of the elevator is upwards

For the block to not slip,
$\Rightarrow m(a+g)\sin {37}^{\circ }\le f_l$
where $f_l=\mu N=\mu m(a+g)cos{37}^{\circ }$

$f_l=0.8\times \frac{4}{5}m(a+g)=\frac{3.2}{5}m(a+g)$

As $\frac{3.2}{5}m(a+g)>m(a+g)sin37$

Therefore, no sliding is possible when the elevator is having upward acceleration. When the elevator is moving upwards, as shown above, the friction is in the upward direction along the incline. So, the displacement and the force are making acute angle which results in a positive work.

Case II; The elevator is moving downwards

For the block to not slip,
Assuming its tendancy to slide downward,
$mg\sin \theta \le f_l+ma\sin \theta$ and
$N=mg\cos \theta -ma\cos \theta$

$f_l=\mu N=0.8m\cos \theta (g-a)=\frac{3.2}{5}m(g-a)$

$\frac{3}{\text{/}5}mg\le \frac{3.2}{\text{/}5}m(g-a)+\frac{3}{\text{/}5}ma$

$3g\le 3.2g-3.2a+3a\Rightarrow 0.2a\le 0.2g$

$\Rightarrow a\le g$
Even if we consider its tendancyt to slide upward , we will get a similar equation of $a\le g$

So, Minimum acceleration of elevator so that block slides over plane is $gm/s^2$