Question

A fixed smooth wedge whose vertical section is a curve y =x^3/3 is shown in the figure.
A small block is released on the curve at location x=1m . The initial acceleration of the block will be

Answer 1

Answer:

$\frac{g}{\sqrt{2}}$

Explanation:

1) We have, fixed smooth wedge whose vertical section is a curved surface with equation as $y=\frac{x^3}{3}$

Now,

We will find angle which the block makes with the horizontal .

Since, block is released at x = 1 m.

Then,

Slope is given by $y'=\frac{dy}{dx}=x^2\\ \\=>y'(1)=1=tan(\theta)\\ \\=>\theta=45^{\circ}$

Hence, the required angle is 45 degree.

2) Now, the acceleration of block is given by

$a=\frac{mgsin(45^{\circ}}{m}=gsin(45^{\circ})=\frac{g}{\sqrt{2}}$