Question

A block of mass 1 kg is at rest relative to a smooth wedge moving leftwards with constant acceleration a=5 ms−2. Let N be the normal reaction between the block and the wedge. Take g=10 ms−2. Choose the correct options from the choices given below:

A
N=55 N
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B
N=15 N
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C
tanθ=12
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D
tanθ=2
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Solution

The correct options are A N=5√5 N C tanθ=12 Since block is at rest w.r.t. wedge, the system of (block + wedge) is moving together with acceleration a leftwards. FBD of Block: Solving from ground frame: Applying equilibrium condition in vertical direction Ncosθ=mg ....(i) Applying Newton's 2nd law in direction of acceleration: Nsinθ=ma ....(ii) Dividing Eq. (ii) and (i) : tanθ=ag=510=12 Putting the value of sinθ from triangle shown in figure, in Eq. (ii): Nsinθ=1×5 ⇒N×1√5=5 ∴N=5√5 N Hence, options A and C are correct.

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