A body starts from rest and slides down an inclined plane(assumed fric...
Given:
Length of the inclined plane, l = 15 m
Time taken to slide down, t = 15 sec
Initial velocity, u = 0 m/s
Acceleration, a = ?
Final velocity, v = ?
Acceleration:
Let's assume that the body slides down the inclined plane with a constant acceleration, a. Using the second equation of motion, we can find the acceleration of the body.
s = ut + 1/2 at^2
l = 0 + 1/2 a(15)^2
a = 2l/t^2
a = 2 x 15 / (15)^2
a = 0.22 m/s^2
So, the acceleration of the body is 0.22 m/s^2.
Final Velocity:
Using the third equation of motion, we can find the final velocity of the body.
v^2 = u^2 + 2as
v^2 = 0 + 2 x 0.22 x 15
v^2 = 6.6
v = √6.6
v = 2.57 m/s
So, the final velocity of the body at the bottom of the inclined plane is 2.57 m/s.
Explanation:
The body starts from rest and slides down an inclined plane. As the inclined plane is assumed to be frictionless, the only force acting on the body is the force due to gravity. Hence, the body slides down the inclined plane with a constant acceleration, a. Using the second equation of motion, we can find the acceleration of the body. Once we find the acceleration, we can use the third equation of motion to find the final velocity of the body at the bottom of the inclined plane.
A body starts from rest and slides down an inclined plane(assumed fric...
S=ut+1/2at^2 (put u=0)..Nd u will get acc. then by relation v^2=u^2+2as(put u=0, Nd a,Nd s,) u will get Ur ans
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