Q37 / 92 · NTA Abhyas (JEE Main) 2020 · Physics · Work Power Energy

A particle of mass m = 1 kg is dropped from a height h = 40 cm on a light horizontal platform fixed to one end of an elastic spring, the other being fixed to a base, as shown in the diagram. The particle collides with the platform and sticks to it. As a result, the spring is compressed by an amount x = 10 cm. What is the force constant of the spring?
(Take g = 10 m s-2)

Figure

  • A
    600 N m-1
     
  • B
    800 N m-1
     
  • Correct answer
    C
    1000 N m-1
     
  • D
    1200 N m-1
     

Solution

Since the platform is depressed by an amount x, the total work done on the spring is mg(h + x). This work is stored in the spring in the form of potential energy 12kx2.

Equating the two, we have

12kx2=mgh+x or  k=2 mgh+xx2

Given, h = 0.4 m, x = 0.1 m, m = 1 kg and g = 10 m s-2. Substituting these values, we get
k = 1000 N m-1.

Hence, the correct answer is option: 'C'.

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