Question

When a person standing on weighing balance, working on principle of Hooke's law, it shows 60kg after a long time and spring gets compressed by 2.5cm, if a person jumps on balance by 10cm , the maximum reading of balance will be?
A) 60kg
B) 120kg
C) 180kg
D) 240kg

Answer 1

Hint: Hooke's law is a law of physics that states that the force (F) needed to extend or compress a spring by some distance (x) scales linearly with respect to that distance—that is, F= Kx , where k is a constant factor characteristic of the spring (i.e., its stiffness), and x is small compared to the total possible deformation of the spring . considered spring force and gravitation force to be conservative and gravitational energy is stored into spring by balancing energy net weight to be calculated.

Complete step by step solution:
Let us make following assumptions
Spring constant = K
Mass when just standing =m kg
Mass when jumping = M kg
Extension in spring = x m
According to first condition as per questions when person is just standing
Weight of person is balanced by spring force
$
kx = mg \\
k \times 2.5 \times {10^{ - 2}} = 60 \times 10 \\
k = 2400N/m \\
\dfrac{1}{2}k{x^2} = mg(0.10 + x) \\
12000{x^2} = 600(0.10 + x) \\
20{x^2} - x - 0.1 = 0 \\
x = 1 \pm \sqrt {1 + 80} \div \left( {2 \times 20} \right) \\
x = 0.1 \\
Mg = kx \\
M \times 10 = 2400 \times 00.1 \\
M = 240kg \\
\\
$
Therefore the weight will be 240kg.

Note: Energy stored in spring is conservative hence in ideal conditions taken to be fixed this whole energy is converted into gravitational potential energy when again one of conservative in nature. Energy just converted in another type.