The initial velocity of a 1.55 kg block sliding down a frictionless inclined plane is found to be 1.27 m/s. Then 1.07 s later, it has a velocity of 5.37 m/s.
What is the angle of the plane with respect to the horizontal?
It's the same question, just put your data..
The initial velocity of a 2.56 kg block sliding down a frictionless inclined plane is found to be 1.13 m/s. Then 1.08 s later, it has a velocity of 3.99 m/s.What is the angle of the plane with respect to the horizontal?
Acceleration = a = (3.99-1.13)/1.08
= 2.648 m/s^2
Force along the plane = M*g sinA = M*a
sinA = a/g = 0.270
A = sin^-1(0.270)= 15.7 degrees
To find the angle of the plane with respect to the horizontal, we can use the principles of motion along an inclined plane.
First, let's define the given information:
- Mass of the block (m) = 1.55 kg
- Initial velocity (v₁) = 1.27 m/s
- Final velocity (v₂) = 5.37 m/s
- Time interval (t) = 1.07 s
To find the angle (θ) of the inclined plane, we can use the equation for velocity along an inclined plane:
v₂ = v₁ + gsinθt
where:
- v₂ = final velocity
- v₁ = initial velocity
- g = acceleration due to gravity
- θ = angle of the inclined plane
- t = time interval
Rearranging the equation, we have:
gsinθ = (v₂ - v₁) / t
Now, let's plug in the values:
gsinθ = (5.37 m/s - 1.27 m/s) / 1.07 s
To solve for gsinθ, we can take the value of acceleration due to gravity (g) as 9.8 m/s²:
9.8 sinθ = (5.37 m/s - 1.27 m/s) / 1.07 s
Now, we can solve for sinθ by dividing both sides of the equation by 9.8:
sinθ = [(5.37 m/s - 1.27 m/s) / 1.07 s] / 9.8
Calculating the right-hand side of the equation, we get:
sinθ = 4.108 / 9.8
Now, we can find the inverse sine (sin⁻¹) of both sides to solve for θ:
θ = sin⁻¹(4.108 / 9.8)
Evaluating this mathematically, we get:
θ ≈ 24.56 degrees
Therefore, the angle of the plane with respect to the horizontal is approximately 24.56 degrees.