badge image

Enroll For Free Now & Improve your performance.

×
User Icon
User Icon
User Icon
User Icon
User Icon

Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
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 man sits on a chair supported by a rope passing over a friction less fixed pulley .The man who weighs 1000N exerts a force of 450N on the chair downwards whule pulling the rope on the other side . If the chair weighs 250N tgen acceleration of chair is?

3 years ago

Answers : (1)

Aman Sharma
40 Points
							
Suppose acceleration is in direction of man going downwards so by drawing the diagram and using fbd consider g to be 10 so the mass of the person is 100kg and 125kg of the chair , we get the following 2 equations 1 T- 450N (force applied by the man) = 45a (450÷g i.e  mg÷mg = m) and 2-   [mg (of man) + mg of chair = (mass of man + mass of chair) so putting values the equation are as follow 1)T - 450N =  45a  , 2) 1250 - T = 125 a.  Solving equation via elimination method we get a= 4.70 
3 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
Vouchers
To Win!!! Click Here for details