A body of mass 2 kg slides down with anacceleration of 3 m/s2 on a rou

1.	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 : (g=10
Answer» ANSWER-6N 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=2101/2=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.

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 : (g=10

Answer»

ANSWER-6N

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.