The velocity of a 2.5 kg block sliding down a frictionless inclined plane is found to be 1.25 m/s. 1.00 s later, it has a velocity of 4.15 m/s.
What is the angle of the plane with respect to the horizontal (in degrees)?
To find the angle of the plane with respect to the horizontal, we can use the equations of motion for an object sliding down an inclined plane.
First, let's consider the initial velocity of the block (v1) and the final velocity of the block (v2). We know that the initial velocity, v1, is 1.25 m/s and the final velocity, v2, is 4.15 m/s.
The equation we can use is:
v2^2 = v1^2 + 2a*d
Where:
v2 is the final velocity
v1 is the initial velocity
a is the acceleration
d is the distance traveled
We also need to consider that the acceleration is due to gravity and is given by:
a = g*sin(theta)
where:
g is the acceleration due to gravity (9.8 m/s^2)
theta is the angle of the inclined plane with respect to the horizontal.
Rearranging the equation, we have:
d = (v2^2 - v1^2)/(2a)
Substituting the value of acceleration, we get:
d = (v2^2 - v1^2)/(2*g*sin(theta))
We can solve for theta by rearranging the equation again:
sin(theta) = (v2^2 - v1^2)/(2*g*d)
Now, substitute the given values into the equation:
sin(theta) = (4.15^2 - 1.25^2)/(2*9.8*d)
We also know that the distance traveled, d, is the product of the initial velocity, v1, and the time, t:
d = v1 * t
Substitute this value for d in the equation:
sin(theta) = (4.15^2 - 1.25^2)/(2*9.8*v1*t)
Now we can calculate the value of sin(theta) using the given values:
sin(theta) = (4.15^2 - 1.25^2)/(2*9.8*1.25*1.00)
Finally, to find the angle theta, we need to take the inverse sine (arcsin) of the calculated value of sin(theta):
theta = arcsin((4.15^2 - 1.25^2)/(2*9.8*1.25*1.00))
Calculating this using a calculator, we find that the angle theta is approximately 33.93 degrees.
Therefore, the angle of the inclined plane with respect to the horizontal is approximately 33.93 degrees.