Question

A bird is sitting in a large closed cage, which is placed on a spring balance. It records a weight of 25 N. The bird (mass=0.5 kg) flies upward in the cage with an acceleration of 2m/$s_2$. The spring balance will now show a record of:
A. 22
B. 23
C. 24
D. 26

Answer 1

Hint: An empty cage should weigh less than when an object is kept in it. But, imagine what a bird does just before flying. Recall Newton’s third law of motion, the bird gets the force to fly by applying the opposite amount of force on the cage.

Complete step-by-step answer:
The second sentence of the question tells that the weight of a bird + cage is 25N. Now, the bird has a mass of 0.5 kg. At this point, the bird jumps with an acceleration of $2\text{m}{{\text{s}}^{-2}}$.
From the second law of motion by newton, which states: “The acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object.”
That is the bird jumps up with a force ${{F}_{u}}$ of:
$\begin{align}
  & \text{Force}=\text{mass}\times \text{acceleration} \\
 & {{F}_{u}}=0.5\times 2=1N \\
\end{align}$
Where does the bird get this force from? The bird pushed the cage downward, which is nothing but manifestation of Newton’s third law of motion which states:
Every action (force) has an equal and opposite reaction(force).
So, the reaction force on the cage is 1N(downward).
So, the weight experienced by the spring balance will be the sum of the recording before and the force due to bird’s motion:
$F=25N+1N=26N$
Therefore, option D is correct.

Note: Firstly, A student might get confused thinking the bird will use its wings to fly and not use its legs for the initial push. But that is not the case. Secondly, after the bird has taken a flight, the spring will read only the mass of the cage which experiences a little velocity.