A block initially at rest, slides down from the top of an inclined plane 30degrees above the horizontal. The plane is 100feet long and frictionless. what distance will it travel on a frictionless horizontal surface at the foot of the plane 6seconds after it has started from rest?
To determine the distance the block will travel on a frictionless horizontal surface at the foot of the plane, we first need to find its velocity when it reaches the bottom of the inclined plane.
We can use the laws of motion to analyze the block's motion. The force acting down the inclined plane is the component of the block's weight parallel to the plane, given by:
Force_parallel = m * g * sin(theta)
where m is the mass of the block, g is the acceleration due to gravity, and theta is the angle of the inclined plane.
Since the inclined plane is frictionless, there are no other forces acting on the block. Therefore, the net force down the plane is equal to the force parallel to the plane:
Net_force = Force_parallel
Using Newton's second law (F = m * a), we can relate the net force to the acceleration:
ma = m * g * sin(theta)
The mass of the block cancels out, giving us:
a = g * sin(theta)
Plug in the values:
a = 9.8 m/s^2 * sin(30°)
a ≈ 4.9 m/s^2
Next, we can find the velocity of the block when it reaches the bottom of the inclined plane using the equation:
v = u + a * t
Since the block starts from rest (u = 0) and the time taken is 6 seconds (t = 6 s), we have:
v = 0 + 4.9 m/s^2 * 6 s
v ≈ 29.4 m/s
Now that we have the velocity, we can determine the distance traveled on the frictionless horizontal surface using the equation:
distance = velocity * time
Given that the time is 6 seconds, the distance traveled is:
distance = 29.4 m/s * 6 s
distance ≈ 176.4 meters
Therefore, the block will travel approximately 176.4 meters on the frictionless horizontal surface at the foot of the inclined plane 6 seconds after starting from rest.