A block of mass m slides on the wooden wedge, which in turn slides backward on the horizontal surface. The acceleration of the block with respect to the wedge is : Given m = 8 kg, M = 16 kg Assume all the surfaces shown in the figure to be frictionless.

Solution

Physics sir · 4 steps

Sit on the wedge: add a pseudo force, find the wedge's acceleration first, and the slide follows.

1. Wedge frame. Let the wedge accelerate left with A. In the wedge's frame the block feels gravity mg, the normal force N and a pseudo force mA pointing right. The block does not move perpendicular to the incline, so N = m(g cos 30° - A sin 30°).
2. Wedge equation. The only horizontal force on the wedge is the horizontal part of N, so N sin 30° = MA. Put in N from above and solve for A.

   A = (mg sin θ cos θ)/(M + m sin² θ) = (8g × 1/2 × √3/2)/(16 + 8 × 1/4) = √3/9 g

3. Along the incline. Down the slope, gravity gives g sin 30° = g/2 and the pseudo force gives A cos 30° = √3/9g × √3/2 = g/6. Both point down the slope, so they add.
4. So. aᵣₑₗ = g/2 + g/6 = 2/3g, which is option (d). The formula aᵣₑₗ = ((M+m)g sin θ)/(M + m sin² θ) = 12g/18 gives the same result.

The idea to keep. For a block on a free wedge, solve the wedge first using only the normal force. Then work in the wedge's frame with a pseudo force. The relative slide comes out more than g sin θ.

Answer from the NTA final answer key. Solution by PrepWiser.