Question

A boy who normally weighs $ 200 N$ on a spring scale suddenly jumps upwards. The scale reading jumps to $400N$. What is boy's maximum acceleration upwards ( $g = 9.8{\text{ m}}{{\text{s}}^{ - 2}}$ )
(A) $2m/{s^2}$
(B) $3m/{s^2}$
(C) $9.8m/{s^2}$
(D) $4.67m/{s^2}$

Answer 1

Hint:To solve this question students must know that an object's apparent weight will differ from an object's real weight whenever the force of gravity acting on the object is not balanced by normal force.


 Complete step by step solution:
Normal weight of the boy on the spring scale$ = 200N$
$g = 9.8{\text{ m}}{{\text{s}}^{ - 2}}$
${\text{Normal weight of the boy = mass }} \times {\text{ gravity}}$
${\text{Normal weight of the boy = mg}}$
Let normal weight of the boy $ = W$
$m = \dfrac{W}{g}$
$m = \dfrac{{200}}{{9.8}}$
$m = 20.40kg$
The equation for when a boy suddenly jumps upwards with a acceleration $a{\text{ }}$
$W'$ be the new weight when the boy jumps upwards $ = 400N$
$W' = mg + ma$
$W' = m\left( {g + a} \right)$
$400N = 20.40\left( {9.8 + a} \right)$
$\dfrac{{400}}{{20.4}} = 9.8 + a$
$19.60 = 9.8 + a$
$a = 9.8m/{s^2}$
Therefore, the correct answer is option (C).

Additional information:
Apparent weight of a body moving in a lift.
When a lift is moving vertically upwards or downwards with uniform velocity then net weight is equal to mg which is equal to the actual weight of the person.
$N = mg$
When a lift is moving vertically upwards with uniform acceleration then
$N = mg + ma$
$N = m(g + a)$
When a lift is moving vertically downwards with uniform acceleration then
$N = mg - ma$
$N = m(g - a)$
When lift falls freely then man feels weightlessness.

Note: Whenever a question is asked of this type, we need to know the concept of force and apparent weight first. Then we are going to consider the situation that is given in the question. Before that, we should do the formula for both the conditions while a body moves with an acceleration an upwards and second condition while a body moves downwards with an acceleration a.