Question

A block of mass m is kept on a wedge of mass 5 kg. Another block of mass m₀ is connected through ideal string and pulley with wedge. If acceleration of block of mass m is only in vertical direction then

• A. Value of m₀ is 15 kg
• B. Value of m₀ is 20 kg
• C. Acceleration of m is 5 m/sec²
• D. Tension in the string is 37.5 N

Answer 1

Answer: (A), (D)

Options A and D are correct.

Answer 2

To have the block of mass m accelerate only vertically, the horizontal component of its acceleration must vanish. This leads to the constraint A + (a_along plane) cos θ = 0 (with A the wedge’s horizontal acceleration). Writing the equations for the block on the wedge (along the plane), for the hanging block, and for the wedge in the horizontal direction yields (after eliminating A)

T = [5m g sin θ/(5+m)] and m₀ = T/g = [5m sin θ/(5+m)].

Consistency then forces m = 75 kg so that

m₀ = 15 kg and T = 37.5 N.

(Also the computed vertical acceleration does not come out as 5 m/s².)