Question

a block of mass m is placed on a smooth triangular block. the triangular block is moving horizontally with uniform speed 2√3 m/s . the acceleration of block of mass and with respect to the triangular block is

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Answer 1

5m/s^2

Explanation:

acceleration of the body on an inclined plane=gsin30=10/2=5m/s^2

since the triangular block is moving with uniform speed there is no acceleration for the triangular block.

therefore relative acceleration of rectangular block with respect to triangular

block= acceleration of block -acceleration of the triangular block

=5-0=5m/s^2

hence 5m/s^2

......Hope its the right ans

Answer 2

Answer:

5 m/s²

Explained answer ☺️

Acceleration of triangular block = 0

(Triangular block is moving with uniform speed)

Acceleration of a block of mass m on a inclined plane=g sin theta

We have angle=30° Then,

=10×1/2=5

Where (g=10 and sin 30°=1/2)

Then we find that

Acceleration of block of mass m with respect to triangular block=acceleration of block of mass m- acceleration of triangular block

We know that by solving above

=5-0

=5 m/s²

I hope this solution is correct