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: 11
        
A suitcase is gently dropped on a conveyor belt moving at 3m/s. If the coefficient of friction between the belt and suitcase is 0.5. Find the displacement of suitcase relative to conveyor belt before the slipping b/w the two is stopped
 
3 years ago

Answers : (1)

Vikas TU
10402 Points
							
Friction Force = umg = mg/2
accln. = > g/2 
No vertical accln. as converyor belt is moving horizontally.
From Newton’s second law of eqn.=>
s = 0.5*a*t^2
 and
0 = u – at
t = u/a => 3/0.5g => 6/g
and distance traveled => 0.5 * g/2 * (6/g)^2 => 0.9 meter approx.
3 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


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

  • 141 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
Vouchers
To Win!!! Click Here for details