a wooden slab starting from rest slide down a 10m long inclined plane with acceleration of 5m per sec what would be the speed ay the bottom of the inclined plane
V^2 = Vo^2 + 2a*d
V^2 = 0 + 10*10 = 100
V = 10 m/s.
To find the speed of the wooden slab at the bottom of the inclined plane, we can use the equations of motion.
First, we need to determine the time it takes for the wooden slab to slide down the inclined plane. To do this, we can use the equation:
s = ut + (1/2)at^2
Where:
- s is the distance traveled (10m in this case)
- u is the initial velocity (0 m/s since it starts from rest)
- a is the acceleration (5 m/s^2)
- t is the time taken (what we are trying to find)
Rearranging the equation to solve for time (t), we get:
10 = (1/2) * 5 * t^2
10 = 2.5t^2
Dividing both sides of the equation by 2.5, we have:
t^2 = 4
Taking the square root of both sides, we find:
t = 2 seconds
Now that we know the time taken, we can find the final velocity (v) using the equation:
v = u + at
Where u is the initial velocity and a is the acceleration. As the wooden slab starts from rest, the initial velocity (u) is 0 m/s. Substituting the values:
v = 0 + 5 * 2
v = 0 + 10
v = 10 m/s
Therefore, the speed of the wooden slab at the bottom of the inclined plane is 10 m/s.