Question 203: Laws of Motion

Question

A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring balance reads $$49 \mathrm{~N}$$, when the lift is stationary. If the lift moves downward with an acceleration of $$5 \mathrm{~m} / \mathrm{s}^2$$, the reading of the spring balance will be $$(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2)$$

Accepted Answer

Here's how to solve this problem:
When the lift is stationary, the spring balance reads the actual weight of the bag, which is $$49 \mathrm{~N}$$. This means the mass of the bag is:
$$m = \frac{W}{g} = \frac{49 \mathrm{~N}}{9.8 \mathrm{~m/s^2}} = 5 \mathrm{~kg}$$
Now, when the lift accelerates downwards, the apparent weight of the bag will change.  Here's why:
The net force acting on the bag is the difference between its weight (mg) and the upward force exerted by the spring balance (let's call it T).  This net force is what causes the bag to accelerate downwards with the lift.
Applying Newton's second law (F = ma):
$$mg - T = ma$$
We can rearrange this to solve for the reading of the spring balance (T):
$$T = mg - ma = m(g - a)$$
Plugging in the values:
$$T = 5 \mathrm{~kg} (9.8 \mathrm{~m/s^2} - 5 \mathrm{~m/s^2}) = 24 \mathrm{~N}$$
Therefore, the reading of the spring balance will be
24 N
when the lift accelerates downwards.
So the correct answer is Option C.

Answer

74 N

Answer

15 N

Answer

24 N

Answer

49 N

Laws of Motion

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