Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

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

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
`        a 263g block is dropped on to a vertical spring with force constant k=2.52N/cm as shown in the below figure the block sticks to the spring ,and the spring compresses 11.8cm before coming momentarily to rest .while the spring is being compressed ,how much work is done (a)by the force of gravity and (b) by the spring © what was the speed of the block just before it hit the spring ?(d)If the initial speed of the block is doubled,what is the maximum compression of the spring?Ignore friction. `
2 years ago

Arun
23492 Points
```							Ke of block = (1/2) m v2 when spring is uncompressed (a) work done by gravity on mass is m g h = .263*9.81*.118 Joules (b) work done by spring on mass is -(1/2)kx2 = -.5(252 N/m)(.118)2 (c) so because v = 0 at bottom initial Ke + work done by g = work done on spring by mass .5 (.263)v2 + .263*9.81*.118 = .5(252 N/m)(.118)2 solve for v
```
2 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on IIT JEE Entrance Exam

View all Questions »

### Course Features

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

### Course Features

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