A block of mass m = 2 kg kept on a wedge of mass M = 9 kg and a horizontal force of 210N is applied on the wedge as shown. All the contacts are smooth and use g = 10m/s 2 . If initially when the force is applied m block is at the bottom of wedge and base length of wedge is 10 m.
(i) Acceleration of wedge when mass m is moving on it –
Text Solution
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(i) :
Let a 2 be acceleration of m w.r.t M
Then,
ma 1 cos θ – mg sin θ = ma 2
– 
= a 2 a 1 – g = a 2 × 
......(i)
N = ma 1 sin θ + mg cos θ ..............(ii)
F– Nsin θ = Ma 1 ...............(iii)
solving above equation solutions are obtained
( ii) :
Let a 2 be acceleration of m w.r.t M
Then,
ma 1 cos θ – mg sin θ = ma 2
– 
= a 2 a 1 – g = a 2 × 
......(i)
N = ma 1 sin θ + mg cos θ ..............(ii)
F– N sin θ = Ma 1 ...............(iii)
solving above equation solutions are obtained
(iii) :
Let a 2 be acceleration of m w.r.t M
Then,
ma 1 cos θ – mg sin θ = ma 2
– 
= a 2 a 1 – g = a 2 × 
......(i)
N = ma 1 sin θ + mg cos θ ..............(ii)
F– Nsin θ = Ma 1 ...............(iii)
solving above equation solutions are obtained
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