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
`        1} When a load of 10 kg is hung from the wire, then extension of 2 m is produced. Then work done by restoring force is_____________?`
one year ago

Arun
22966 Points
```							F = -kx is the restoring force k is the spring constant => load of 10 kg weighs 10g N => 10(9.81) = k(2) => k = 49.05 N/m therefore work done = forceₐᵥₑ x displacement in direction of force here forceₐᵥₑ(kx/2) and distance moved are oppositely directed W =[ -49.05 + 0] *2*2/2 = - 98.1 J this work done is stored in the wire as elastic potential energy(½kx²)
```
one year ago
Think You Can Provide A Better Answer ?

## Other Related Questions on General Physics

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

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