Laws of MotionMedium
Question
Figure shows a wedge of mass 2 kg resting on a frictionless floor. A block of mass 1 kg is kept on the wedge and the wedge is given an acceleration of 5 m/sec2 towards right. Then :
Options
A.block will remain stationary w.r.t. wedge
B. the block will have an acceleration of 1 m/sec2 w.r.t. the wedge
C.normal reaction on the block is 11 N
D.net force acting on the wedge is 2 N
Solution
FBD of block is shown w.r.t. wedge and FBD of wedge is shown w.r.t. ground. FP is pseudo force.
mg sin 37 − ma cos 37 = mab
⇒ ab = g sin 37 − a cos 37
⇒ N = 1 × 10 × 4/5 + 1 × 5 ∉ 3/5
N = 11 N.
Net force acting on block w.r.t. ground
F =
=
=
F = √ N
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