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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 : (
A body of mass 2 kg slides down with anacceleration of 3 m/s2 on a rough inclinedplane having a slope of 30. The externalforce required to take the same body upthe plane with the same acceleration willbe : (

```
2 years ago

Arun
25768 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
```
2 years ago
john cena
35 Points
```							hence acording to angle of repose, tan 60=coefficient of frictionlet us say coefficient of friction=K=tan 60so friction(f) apply force in upward direction as (f) = Kmgcos30 on resolving mg component we get, mgsin30 in downward direction and Kcos30 in upward directionhence 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 forceso 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^2So F + Kmgcos30 − mgsin30 = ma      putting values      F + 10 − 10 = 2 * 3      F = 6 NHence required force to get an acceleration of 3m/s^2 in upward drection is 6 N.
```
2 years 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/s210 sin30-k10cos30=3 =>  10x1/2-kx10xroot3/2=3  =>5-k5root3=3 =>k=2/5root3Therefore the  force reqd  push up the body with the same acceleration ismgsinx+kmgcosx=2x10X1/2+2/5root3 x2x10xroot3/2=10+4=14 N  ( root3= 1.732, cos30=root3/2 and sin30=1/2ans: 14Nkkbisht
```
2 years ago
john cena
35 Points
```							 hence acording to angle of repose, tan30=coefficient of frictionlet us say coefficient of friction=K=tan 30so friction(f) apply force in upward direction as (f) = Kmgcos30 on resolving mg component we get, mgsin30 in downward direction and Kcos30 in upward directionhence 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 forceso 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^2So F + Kmgcos30 − mgsin30 = ma      putting values      F + 10 − 10 = 2 * 3      F = 6 NHence required force to get an acceleration of 3m/s^2 in upward drection is 6 N.
```
2 years ago
ashmitha
15 Points
```							hence acording to angle of repose,tan 60=coefficient of friction=root 3let us say coefficient of friction= K=tan 60so friction(f) apply force in upward direction as (f) = Kmgcos30 on resolving mg component we get, mgsin30 in downward direction and Kcos30 in upward directionhence 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 forceso 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^2So F + Kmgcos30 − mgsin30 = ma      putting values      F + 10 − 10 = 2 * 3      F = 6 NHence required force to get an acceleration of 3m/s^2 in upward drection is 6 N.
```
2 years ago
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