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 body of mass 40 kg stands on a weighing machine in a accelerated lift. The reading on the scale of the weighing machine is 300N. Find the magnitude and direction of acceleration. (g=9.8 m/s^2)
2 years ago

Answers : (1)

Arun
23506 Points
							
Dear student
 
M=40kg
F=300N
suppose that g=9.8m/s²

Now, when the lift goes down, 2 types of accelerations are applied, 
1. gravitational acceleration
2.acceleration of life

By thinking about the FBD of the lift,
if the lift moves down, the sum of acceleration = g-a
and if it moves up, it is g+a

[why? Becuase when the life moves, an illusory acceleration is applied on it, which is 'a' here]

That means, 
1. if the lift is moving upwards,
F=m(g+a)
300=40(10+a)
a= -2/5 m/s²

2. if lift is going down,
F=m(g-a)
300=40(10-a)
a= +2/5 m/s²

Now the directions:
 1.if the lift moves downwards, the magnitude of acceleration is +2/5 m/s²
2.if the lift moves downwards, the magnitude of acceleration is -2/5 m/s²
 
Regards
Arun (askIITians forum expert)
2 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