The velocity of a 2.3 kg block sliding down a frictionless inclined plane is found to be 1.58 m/s. 1.1 s later, it has a velocity of 5.65 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 formula for the acceleration of an object sliding down an inclined plane.
The formula is given as: a = g * sin(θ), where a is the acceleration, g is the acceleration due to gravity, and θ is the angle of the plane with respect to the horizontal.
Given that the block is sliding down a frictionless inclined plane, we know that there is no force opposing the motion, so the acceleration is constant. Therefore, the change in velocity (Δv) of the block over the time interval of 1.1 seconds is equal to the acceleration multiplied by the change in time (Δt).
Δv = a * Δt
Substituting the values given into the equation, we have:
5.65 m/s - 1.58 m/s = a * (1.1 s)
Simplifying the equation, we have:
3.82 m/s = a * 1.1 s
Now, we need to find the acceleration a. Using the formula for the acceleration of an object sliding down an inclined plane, a = g * sin(θ), we can rearrange the equation to solve for the angle θ.
θ = arcsin(a/g)
Substituting the values of a and g into the equation, we have:
θ = arcsin((3.82 m/s) / (9.8 m/s^2))
Calculating the value of arcsin using a calculator, we find that:
θ ≈ 23.8 degrees
Therefore, the angle of the plane with respect to the horizontal is approximately 23.8 degrees.