Question

a body is released along a smooth inclined plane of inclination of 30 ,and lemgth is 40 m.find the time taken to reach the bottom​

Answer 1

Answer

• Time taken to reach bottom is 4 seconds

Given

• a body is released along a smooth inclined plane of inclination of 30 ,and length is 40 m

To Find

• Time taken to reach the bottom

Concept Used

Find attachment for diagrammatic view

Slant height = l

Initial velocity , u = 0 m/s [ ∵ released from rest ]

Acceleration , a = g sinθ

Time = t

Apply 3rd equation of motion .

$\implies \rm s=ut+\dfrac{1}{2}at^2\\\\\implies \rm l=(0)t+\dfrac{1}{2}(gsin \theta)t^2\\\\\implies \rm l=\dfrac{g\ sin\theta .t^2}{2}\\\\\implies \rm t^2=\dfrac{2l}{g\sin \theta}\\\\\implies \bf t=\sqrt{\dfrac{2l}{g.\sin \theta}}$

This is total time that is taken to reach bottom

Solution

Angle of inclination , θ = 30°

Length , l = 40 m

Acceleration , a = g = 10 m/s²

Time , t = ? s

Apply formula .

$\implies \rm t=\sqrt{\dfrac{2l}{g.\sin \theta}}\\\\\implies \rm t=\sqrt{\dfrac{2(40)}{10.sin30^0}}\\\\\implies \rm t=\sqrt{\dfrac{80}{10 \times \frac{1}{2}}}\\\\\implies \rm t=\sqrt{\dfrac{80}{5}}\\\\\implies \rm t=\sqrt{16}\\\\\implies \rm t=4\ s$