Question

A block is placed on top of a fixed smooth inclined plane inclined at 30° to horizontal. The length of plane is 5 m. The block slides down the plane and reaches at bottom. The speed of the block at bottom will be nearly
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Answer 1

Answer:

the correct option is 2

the answer is 7

Explanation:

v2= u2 +2as

v2= 02 + 2 × gsintheta × s

v= whole under root of

2 x 9.8 x 1/2 x 5 = √49 = 7

Answer 2

The speed of the block will be 7m/s.

Step by Step explanation:

Given: Inclined plane angle= 30°

            Length of plane (l) = 5m

To be found: Speed of block at bottom

Solution:

Let us assume height of plane as 'h'

∴ according to trigonometry:

$sin 30 = \frac{h}{5}$

⇒$\frac{1}{2}= \frac{h}{5}$

⇒$h= 2.5m$

We know, when an object slides on an inclined plane:

Loss of Potential energy (PE) = Gain of kinetic energy (KE)

∴ $mgh= KE_{1}-KE_{2}$

As the block is placed at a height, there is no initial kinetic energy, so $KE_{1} = 0$

∴$mgh= \frac{1}{2}mv^{2} - 0$

⇒$gh= \frac{1}{2} v^{2}$

⇒$v= \sqrt{2gh}$

putting values of g= $9.8m/s^{2}$ and h=$2.5m$

we get v= 7 m/s.

Hence, The speed of the block at bottom will be 7m/s.