The velocity of a 1.5 kg block sliding down a frictionless inclined plane is found to be 1.13 m/s. 1.2 s later, it has a velocity of 4.37 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 concepts of acceleration due to gravity and kinematic equations.
Let's start by finding the acceleration of the block. We know that the block is sliding down a frictionless inclined plane, so the only force acting on it is the force due to gravity. The acceleration of the block can be calculated using the formula:
a = g * sin(theta)
where:
a is the acceleration of the block,
g is the acceleration due to gravity (approximately 9.8 m/s^2),
theta is the angle of the plane with respect to the horizontal.
Next, we can use the kinematic equation to find the displacement of the block during the 1.2 second interval:
v = u + at
where:
v is the final velocity (4.37 m/s),
u is the initial velocity (1.13 m/s),
a is the acceleration of the block (which we calculated earlier),
t is the time interval (1.2 seconds).
Rearranging the equation, we have:
a = (v - u) / t
Substituting the given values, we can calculate the acceleration of the block.
Finally, once we have the acceleration, we can use the inverse sine function to find the angle of the plane:
theta = arcsin(a / g)
Substituting the known values, we can calculate the angle of the plane with respect to the horizontal.
Let's plug in the numbers and calculate it.
To find the angle of the plane with respect to the horizontal, we can use the kinematic equations of motion.
Let's start by finding the acceleration of the block.
We can use the equation:
v = u + at,
where:
v = final velocity = 4.37 m/s,
u = initial velocity = 1.13 m/s,
a = acceleration,
t = time interval = 1.2 s.
Rearranging the equation to solve for a, we get:
a = (v - u) / t.
Substituting the given values, we have:
a = (4.37 m/s - 1.13 m/s) / 1.2 s.
a = 3.24 m/s / 1.2 s.
a = 2.7 m/s².
Now, we can find the angle using the equation:
a = g * sin(θ),
where:
g = acceleration due to gravity = 9.8 m/s²,
θ = angle of the plane.
Rearranging the equation to solve for θ, we have:
θ = arcsin(a / g).
Substituting the known values, we get:
θ = arcsin(2.7 m/s² / 9.8 m/s²).
Using a calculator, we find:
θ ≈ 0.2776 radians.
To convert radians to degrees, we multiply by 180/π:
θ ≈ 0.2776 * 180/π ≈ 15.91 degrees.
Therefore, the angle of the plane with respect to the horizontal is approximately 15.91 degrees.
It is accelerating at a rate (4.37-1.13)/1.2 = 2.7 m/s^2
The Force pushing it down the ramp is
F = a/M = 1.8 N
With no friction, 1.8 N must be the component of the weight down the ramp, which is M g sin A. Therefore
1.8 N = (1.5)(9.8) sinA
sin A = 0.1224
A = 7.0 degrees