Question

29. A block sliding along a horizontal frictionless

surface with a velocity of 2 m s' comes to the

bottom of a frictionless inclined plane making

an angle of 30° with the horizontal and comes

to a stop after ascending the inclined plane as

indicated. If g = 10 ms, the value of 'h' is

stops here

2 m/s

1) 5 m

2) 10

m

3) 0.2 m

4) 2 m

.

Answer 1

Answer:

option (c) is correct

plz mark me as brainiliest

Answer 2

Answer:

The correct option for the value of height 'h' of a frictionless inclined plane in which a block slides and comes to rest at the bottom of the incline is option 'c' 0.2m

Explanation:

In the case of a block sliding in an inclined plane, the acceleration that acts in the direction of motion is  $gsin\theta$

Step 1:

Given initial velocity u=2m/s, g=10m/$s^{2}$ and the $\theta=30^{\circ}$. Since the block comes to a stop, final velocity v=0m/s

The component along the direction of motion of the block is

                                        $gsin\theta\\gsin(30^{\circ})=g/2$

Step 2:

Using the Equation of motion

                                       $v^{2}-u^{2}=2as$  - ---------(1)

where 'a' is acceleration and 's' is the displacement of the particle.

Step 3:

In our case                    $a=gsin\theta=g/2$

Substituting the values in equation (1) we get,

                                     $2^{2}-0^{2}=\frac{2gs}{2}\\ 4=gs$---------------(2)

Step 4:

The displacement s is the hypotenuse of the incline. We can write s in terms of $\theta$ and h.

                                    $sin\theta=\frac{opposite side}{hypotenuse}\\sin\theta=\frac{h}{s}$

Rearranging we get,

                                    $s=\frac{h}{sin\theta}$

Step 5:

Substituting for s in equation (2),

                                   $4=\frac{gh}{sin\theta}\\h=\frac{4sin\theta}{g}\\ \theta=30^{\circ}\\sin30^{\circ}=1/2\\h=\frac{4}{2g} \\h=0.2m$

Therefore, the height of the incline for which a particle sliding along the incline comes to rest after ascending is 0.2m