A block slides down a frictionless plane having an inclination of 18.9deg. If the block starts from rest at the top and the length of this incline is 2.21m, find the acceleration of the block.
b). What is the block's speed when it reaches the bottom of the incline?
To find the acceleration of the block sliding down the frictionless plane, we can use the equations of motion.
First, let's find the acceleration. The acceleration of an object sliding down a frictionless plane is equal to the component of gravitational force acting parallel to the incline. This can be calculated using the formula:
a = g * sin(theta)
where a is the acceleration, g is the acceleration due to gravity (approximately 9.8 m/s^2), and theta is the angle of inclination.
Plugging in the values given, we have:
a = 9.8 m/s^2 * sin(18.9 deg)
Calculating this, we find:
a ≈ 3.16 m/s^2
So, the acceleration of the block sliding down the incline is approximately 3.16 m/s^2.
To find the speed of the block when it reaches the bottom of the incline, we can use the equation of motion:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity (which is 0 because the block starts from rest), a is the acceleration, and s is the distance traveled down the incline.
Plugging in the values given, we have:
v^2 = 0 + 2 * 3.16 m/s^2 * 2.21 m
Calculating this, we find:
v^2 ≈ 13.94 m^2/s^2
Taking the square root of both sides, we get:
v ≈ √13.94 m/s
So, the block's speed when it reaches the bottom of the incline is approximately 3.73 m/s.
To find the acceleration of the block, we can use the formula for the acceleration of a block on an inclined plane:
Acceleration (a) = g * sin(theta)
where g is the acceleration due to gravity (9.8 m/s^2) and theta is the angle of incline (18.9 degrees).
Substituting the values into the formula:
a = 9.8 * sin(18.9)
Using a calculator, we find:
a ≈ 3.27 m/s^2
So, the acceleration of the block is approximately 3.27 m/s^2.
To find the block's speed when it reaches the bottom of the incline, we can use the equation relating speed, acceleration, and distance:
v^2 = u^2 + 2ad
where v is the final velocity, u is the initial velocity (which is zero in this case since the block starts from rest), a is the acceleration, and d is the distance traveled (which is 2.21 m in this case).
Substituting the values into the equation:
v^2 = 0^2 + 2 * 3.27 * 2.21
v^2 ≈ 14.4
Taking the square root of both sides, we find:
v ≈ 3.79 m/s
So, the block's speed when it reaches the bottom of the incline is approximately 3.79 m/s.