When a stone with mass 3 kg falls from a height of 10 m on a spring with spring constant 100 N/m, by how much is the spring compressed.

Expert Answers

An illustration of the letter 'A' in a speech bubbles

The stone with mass 3 kg is initially at a height of 10 m. It is dropped on a spring with spring constant what 100 N/m. The length by which the spring is compressed has to be determined.

At a height of 10 m, the gravitational potential energy of the stone is PE = m*g*h = 3*9.8*10 = 294 J. When a spring is compressed by x m, the potential energy in the spring is given by (1/2)*k*x^2 where k is the spring constant. Here, k = 100. It is assumed that the entire energy in the stone is stored as potential energy in the spring.

If the spring is compressed by x m, `(1/2)*100*x^2 = 294`

=> `x^2 = 294/50`

=> `x ~~ 2.42` m

The spring is compressed by 2.42 m

Approved by eNotes Editorial Team

We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

  • 30,000+ book summaries
  • 20% study tools discount
  • Ad-free content
  • PDF downloads
  • 300,000+ answers
  • 5-star customer support
Start your 48-Hour Free Trial