The velocity of a 2.5 kg block sliding down a frictionless inclined plane is found to be 1.58 m/s. 1.40 s later, it has a velocity of 7.32 m/s.
What is the angle of the plane with respect to the horizontal (in degrees)?
acceleration= change velocity/time
force down= mass*acceleration
mgSinTheta= m*a
solve for theta.
To find the angle of the plane with respect to the horizontal, we can apply the principles of kinematics and trigonometry.
First, let's analyze the motion of the block on the inclined plane. We have two sets of velocity values: 1.58 m/s at time t=0 and 7.32 m/s at time t=1.40 s.
Using these velocity values, we can find the acceleration (a) of the block on the inclined plane using the equation:
a = (final velocity - initial velocity) / time
a = (7.32 m/s - 1.58 m/s) / (1.40 s - 0 s)
= 5.74 m/s / 1.4 s
= 4.1 m/s^2
The acceleration represents the component of gravity acting parallel to the inclined plane.
Now, we can find the angle of the plane (θ) using the equation:
sin(θ) = a / g
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Rearranging the equation, we have:
θ = arcsin(a / g)
θ = arcsin(4.1 m/s^2 / 9.8 m/s^2)
= arcsin(0.42)
Using a scientific calculator, we can solve for θ by taking the arcsin of 0.42. The result is approximately 24.6 degrees.
Therefore, the angle of the plane with respect to the horizontal is approximately 24.6 degrees.