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:
        
A body of mass 2 kg slides down with an
acceleration of 3 m/s2 on a rough inclined
plane having a slope of 30. The external
force required to take the same body up
the plane with the same acceleration will
be : (
one year ago

Answers : (5)

Arun
22998 Points
							
Contact force is the resultant of Frictional force and normal reaction force. The frictional force can be calculated as = 0.5 x mgcos theta. = 0.5 x 2 x 10 x root 3 / 2 normal reaction force is Mgcos theta. = 2 x 10 x root 3 / 2
one year ago
john cena
35 Points
							
hence acording to angle of repose, tan 60=coefficient of friction
let us say coefficient of friction=K=tan 60
so friction(f) apply force in upward direction as (f) = Kmgcos30 
on resolving mg component we get, mgsin30 in downward direction and Kcos30 in upward direction
hence we want acceleration in upward direction so there will be a need of external force in upward direction let us say F be the external force
so K=tan30=1/1.73 N
      (f)= Kmgcos30=1/1.73 * 1.73/2 * 20 = 10 N
     mgsin30=2*10*1/2=10 N
                 m=2kg
                   a=3m/s^2
So F + Kmgcos30 − mgsin30 = ma
      putting values
      F + 10 − 10 = 2 * 3
      F = 6 N
Hence required force to get an acceleration of 3m/s^2 in upward drection is 6 N.
one year ago
kkbisht
90 Points
							
The equation of motion of the body sliding down the inclined plane of slope 30 degree is:
mgsinx-kmgcosx=ma  ( here x is slope of the inclined plane and k= coefficient of friction)
put m= 2kg, a=3m/s2,       x=30 degree, g=10 m/s2
10 sin30-k10cos30=3 =>  10x1/2-kx10xroot3/2=3  
=>5-k5root3=3 =>k=2/5root3
Therefore the  force reqd  push up the body with the same acceleration is
mgsinx+kmgcosx=2x10X1/2+2/5root3 x2x10xroot3/2=10+4=14 N  ( root3= 1.732, cos30=root3/2 and sin30=1/2
ans: 14N
kkbisht
 
 
one year ago
john cena
35 Points
							
 
hence acording to angle of repose, tan30=coefficient of friction
let us say coefficient of friction=K=tan 30
so friction(f) apply force in upward direction as (f) = Kmgcos30 
on resolving mg component we get, mgsin30 in downward direction and Kcos30 in upward direction
hence we want acceleration in upward direction so there will be a need of external force in upward direction let us say F be the external force
so K=tan30=1/1.73 N
      (f)= Kmgcos30=1/1.73 * 1.73/2 * 20 = 10 N
     mgsin30=2*10*1/2=10 N
                 m=2kg
                   a=3m/s^2
So F + Kmgcos30 − mgsin30 = ma
      putting values
      F + 10 − 10 = 2 * 3
      F = 6 N
Hence required force to get an acceleration of 3m/s^2 in upward drection is 6 N.
one year ago
ashmitha
15 Points
							
hence acording to angle of repose,
tan 60=coefficient of friction=root 3
let us say coefficient of friction= K=tan 60
so friction(f) apply force in upward direction as (f) = Kmgcos30 
on resolving mg component we get, mgsin30 in downward direction and Kcos30 in upward direction
hence we want acceleration in upward direction so there will be a need of external force in upward direction let us say F be the external force
so K=tan30=1/1.73 N
      (f)= Kmgcos30=1/1.73 * 1.73/2 * 20 = 10 N
     mgsin30=2*10*1/2=10 N
                 m=2kg
                   a=3m/s^2
So F + Kmgcos30 − mgsin30 = ma
      putting values
      F + 10 − 10 = 2 * 3
      F = 6 N
Hence required force to get an acceleration of 3m/s^2 in upward drection is 6 N.
11 months 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