Question

A student is skateboarding down a ramp that is 6.0 m long and inclined at 180 with respect to the horizontal. The initial speed of the skateboarder at the top of the ramp is 2.6 m/s. Neglect friction and find the speed at the bottom of the ramp.

Answer 1

Answer:

The speed at the bottom of the ramp is 2.6m/s

Explanation:

The final velocity can be determined by means of the equations for a Uniformly Accelerated Rectilinear Motion:

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

$v_{f} = \sqrt{v_{i}^{2} + 2ad}$ (1)

Notice that it is necessary to found the acceleration that can be done by means of Newton's second law:

The acceleration can be found by means of Newton's second law:

$\sum F_{net} = ma$

Where $\sum F_{net}$ is the net force, m is the mass and a is the acceleration.

$Fx + Fy = ma$ (2)

Force in the x axis:

$F_{x} = W_{x}$  

The component of the weight in the x axis can be gotten by means of trigonometric:

$\frac{OC}{H} = sen \theta$

$\frac{W_{x}}{W} = sen \theta$

$W_{x} = W sen \theta$

$W_{x} = mgsen \theta$

$F_{x} = mgsen \theta$  (3)

Forces in the y axis:

$F_{y} = N - W_{y}$ (4)

The component of the weight in the y axis can be gotten by means of trigonometric:

$\frac{AC}{H} = cos \theta$

$\frac{W_{y}}{W} = cos \theta$

$W_{y}= W cos \theta$

Remember that the weight is defined as:

$W = mg$

$W_{y}= mgcos \theta$

The normal force can be obtained from equation (4)

$N - W_{y} = 0$

$N = W_{y}$

$N = mgcos \theta$

Therefore, equation 4 can be rewritten as:

$F_{y} = mgcos \theta - mgcos \theta$

$F_{y} = 0$ (5)

Then, replacing equation 3 and equation 5 in equation 2 it is gotten:

$mgsen \theta + 0 = ma$

$mgsen \theta = ma$ (6)

However, a can be isolated from equation 6

$a = \frac{mgsen \theta}{m}$

$a = gsen \theta$  (7)

Finally, equation 7 can be replaced in equation 2:

$v_{f} = \sqrt{v_{i}^{2} + 2d(gsen \theta)}$

$v_{f} = \sqrt{(2.6m/s)^{2} + 2(6.0m)(9.8m/s^{2})sen 180^{\circ})}$

$v_{f} = 2.6m/s$

Hence, the speed at the bottom of the ramp is 2.6m/s