Question

A 6.00-kg block starts from rest and slides down a frictionless incline. When the block has slid a distance 2.00 m, its speed is 3.00 m/s. At what angle above horizontal is the inclined plane tilted?

Answer 1

Answer:

The angle above horizontal is 13.3°.

Explanation:

The angle can be calculated with Newton's third law:

$\Sigma F = ma$

Where:

ΣF: is the forces acting on the object

m: is the mass of the object = 6.00 kg

a: is the acceleration of the object

The only force acting on the object is the weight since there is no friction, so:                  

$mgsin(\theta) = ma$

$gsin(\theta) = a$   (1)

Where:

θ: is the angle

g: is the acceleration due to gravity = 9.81 m/s²

We can find the acceleration from the following kinematic equation:

$v_{f}^{2} = v_{0}^{2} + 2ad$

Where:

$v_{f}$: is the final speed = 3.00 m/s

$v_{0}$: is the initial speed = 0 (the block starts from rest)

d: is the distance traveled = 2.00 m

The acceleration is:

$a = \frac{v_{f}^{2}}{2d} = \frac{(3.00 m/s)^{2}}{2*2.00 m} = 2.25 m/s^{2}$

Finally, the angle is (equation 1):

$\theta = sin^{-1}(\frac{a}{g}) = sin^{-1}(\frac{2.25 m/s^{2}}{9.81 m/s^{2}}) = 13.3$

Therefore, the angle above horizontal is 13.3°.

I hope it helps you!