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
Menu
Grade: 9
        
A BLOCK OF MASS 2 KG IS PLACED ON THE FLOOR. THE COEFFICIENT OF STATIC FRICTION IS 0.4. IF  A FORCE OF 1.25 ACTS FOR 8 SEC. FINAL VELOCITY OF THE BODY IS
10 months ago

Answers : (1)

Arun
14829 Points
							

We have, coefficient of static friction, µs = 0.4

Mass, m = 2 kg

Force applied, F = 2.5 N

Maximum static frictional force, fmax= µs × m × g = 0.4 × 2 × 9.8

=> fmax = 7.84 N

Since, the applied force is less than the maximum static frictional force, the frictional force on the block is equal to the applied force = 2.5 N. This is according to the fact that static friction is a self adjusting force.

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

Get Extra Rs. 1,590 off

COUPON CODE: SELF10


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


Get Extra Rs. 64 off

COUPON CODE: SELF10

Ask Experts

Have any Question? Ask Experts

Post Question

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