Box on top of a plane!

Classical Mechanics Level 4

A block of mass $5 \text{kg}$ is placed on top of an aeroplane moving horizontally with an acceleration of $11 {\text{m/s}}^2$ at a height of $500 \text{m}$ above the ground. If the box is placed at a distance of $10 \text{m}$ from the tail of the aeroplane, then calculate the total time taken by the block to reach the ground.

Details and assumptions :-

Assume that the air resistance is negligible.
Assume that the aeroplane is of negligible width and does not change its direction during flight.
Assume that the box does not collide with the tail of the aeroplane just before falling off the aeroplane.
Take ${\mu}_{\text{s}}$ (co-efficient of static friction) between the block and the aeroplane's surface as $0.5$ .
Take $\text{g}$ (acceleration due to gravity) as $10 {\text{m/s}}^2$ .
Give your answer (in seconds) correct to two decimal places.

The answer is 11.83.

2 solutions

Ashish Menon
Jul 19, 2016

Frictional force acting on the block = ${\mu}_s × R\\ = 0.5 × 5 × 10\\ = 25N$

Force applied on the block by forward velocity of aeroplane = $m×a\\ = 5 × 11\\ = 55N$

So, the net force applied = $55 - 25\\ = 30N$

So, relative force acting on the block = $m×a$ $30 = 5a\\ a = 6 {\text{m/s}}^2$

So, the time at which the block falls:-
$S = ut + \dfrac{1}{2} at^2\\ 10 = 0×t + \dfrac{1}{2} × 6t^2\\ t^2 = \dfrac{10}{3}\\ t = \sqrt{\dfrac{10}{3}} \ \text{sec}$

Now after falling down, the time taken to reach the ground is given by $\sqrt{\dfrac{2h}{g}}\\ = \sqrt{\dfrac{2×500}{10}}\\ = 10 \ \text{sec}$

So, total time taken by the block to fall down = $10 + \sqrt{\dfrac{10}{3}}\\ \approx \color{#3D99F6}{\boxed{11.83 \ \text{sec}}}$

You keep mixing up between block and box...which is it?

Anyway, for better alignment, include the first "=" in LaTeX:

Frictional force acting on the box $\color{#3D99F6}{=}\mu_s\times R\\ =0.5\times 5 \times 10\\ =25N$

Typo: $=m \times a\\ 30=5a\\ a=6 \text{m s}^{-2}$

Hung Woei Neoh - 4 years, 11 months ago

Hehe that box and blocks does not make much difference I guess. :P And yeah I know that trick but I wrote this solution in a hurry. I dont get much time nowadays :/

Ashish Menon - 4 years, 10 months ago

Neither do I. I have a list of "future questions to post", which I will only post when I have time, since the solutions would probably take at least half an hour to write

Hung Woei Neoh - 4 years, 10 months ago

@Hung Woei Neoh – I am inclined towards classical mechanics section now (you must have observed) so they are random questions. I dont have a certain list. :P

Ashish Menon - 4 years, 10 months ago

The question is ambiguous - it does not indicate the direction of acceleration.

Eric Chan - 4 years, 10 months ago

Why is that, if you think that the block will shift towards the head of the plane and fall, then you are wrong, cuz whichever direction the plane flies, the block will travel towards the tail and fall off simce it is moving wiyh an acceleration and not retardation. Again, if you think it flies in an opposite direction, no worries, it can fly in that direction, the question just asks the time taken to reach the ground which is independent of the direction the aeroplane travels.

Ashish Menon - 4 years, 10 months ago

Logically, I can assume the acceleration (which you did not indicate the direction) having a vertical component of 10 m/s^2 to balance the gravity and having an unknown horizontal component to be solved.

The direction of the plane flying is of course significant, it can be acceleration vertically upwards , in which the time will be infinite. I just gave you an example to show the necessity of clarifying all vectors' direction in a proper physics question.

Instead of using common sense, a rigorous definition of physical bodies ' kinetic condition have to be well - defined. In every case, I will assume ''aeroplane" as a name for a physical block, as you do not specify anything about its physical status beside 500m from the ground and 11 m/s^2 acceleration in an unknown direction.

Eric Chan - 4 years, 10 months ago

@Eric Chan – Aa I see what you mean, its obvious but I will add that in. And by the way if it was moving upwards, then the statement that the plane is at a height of 500m above the ground will not be stated in the first place. So, we can deduce (not assume) that it is moving horizontally.

Ashish Menon - 4 years, 10 months ago

@Ashish Menon – Just for doing physics I will challenge every blindspot :) no offence and I just switch from the maths sector to here.

You are right that 500 m implication is not catching my attention.

Eric Chan - 4 years, 10 months ago

@Eric Chan – :P lel, yeah I am interested in physocs nowadays too.

Ashish Menon - 4 years, 10 months ago

Sargam Yadav
Jul 20, 2016

Frictional force = 5×10×0.5=25N Force due to acceleration of aeroplane =5×11=55N Net force on the block =55-25=30N So,net acceleration = 30/5=6m/s^2. Now, s=1/2( at^2). So,10=1/2 (6t^2). therefore, t= √(10/3)sec.( Which is time for covering 10 m on the aeroplane ). Now,time t taken by box to come down : 500=1/2(10 t^2).. So,t= 10 sec Total time taken = 10+√(10/3) = 11.83

But when the block falls down its horizontal projectile ..?? Am I right

Prabhat Rao - 4 years, 10 months ago

Yes,you are correct.. you can do by that also .. as the formula will be same .I.e root 2 h/ g.

Sargam yadav - 4 years, 10 months ago