Question

In the figure shown all the surfaces are frictionless, and mass of the block, m= 1 kg. The block and
wedge are held initially at rest. Now wedge is given a horizontal acceleration of 10 m/s by applying
a force on the wedge, so that the block does not slip on the wedge. Then work done by the normal
force in ground frame on the block in
$\sqrt{3}$
seconds is :
10m/s​

Answer 1

Answer:

150 J

Explanation:

1) Let the Normal Reaction of Wedge on the Block be 'N'.

We know,

Acceleration of Block , a = 10 m/s^2

Initial velocity of block , u = 0 m/s

Using Newton's Equation of motion,

$t=\sqrt{3}s\\ S=ut+\frac{1}{2}at^2\\ \\ =0+\frac{1}{2}*10*(\sqrt{3})^2\\ \\=15 m$

2) Now,

Using Newton's First law,

$Nsin(\theta)=ma = 1*10=10$

3) Work done by the Normal Force,

$W_N=N*S*cos(\frac{\pi}{2}-\theta)\\ \\=N*15*sin(\theta)\\ \\=Nsin(\theta)*15\\ \\=10*15\\ \\ =150 N$

Hence, Work Done by the Normal Force in ground Frame on the block is 150 N.

Answer 2

Answer:

150 absolutely correct!

Explanation: