Use Coupon: CART20 and get 20% off on all online Study Material

Total Price: Rs.

There are no items in this cart.
Continue Shopping
Grade: 12


The system is released frm rest with the spring initially stretched 75 mm. The spring has a stiffness of 1050 N/m. Show that the velocity of the block after it has dropped 12 mm will be 0.371 m/s. Neglect the mass of the small pulley

10 years ago

Answers : (2)

Sanchit Gupta
8 Points

NOTE the mass of the block is 45 kg.

10 years ago
Badiuddin askIITians.ismu Expert
147 Points

HI sanchit gupta

Here one thing shoul be noted that when block falls 12 mm then spring must be streched by 2*12=24 mm


now energy balance

initial energy = 45*g*.012 +(1/2) *1050 *(.075)2

final energ   = (1/2) *45 *v2 +(1/2) *1050*(.075+.024)2


equate these two and calculate v

v=.37 m/s

Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly.
 We are all IITians and here to help you in your IIT JEE preparation.

 All the best Sanchit.
Askiitians Experts

10 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

Course Features

  • 101 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution

Course Features

  • 110 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions

Ask Experts

Have any Question? Ask Experts

Post Question

Answer ‘n’ Earn
Attractive Gift
To Win!!! Click Here for details