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:
[tex]v_{f}^{2} = v_{i}^{2} + 2ad[/tex]
[tex]v_{f} = \sqrt{v_{i}^{2} + 2ad}[/tex] (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:
[tex]\sum F_{net} = ma[/tex]
Where [tex]\sum F_{net}[/tex] is the net force, m is the mass and a is the acceleration.
[tex]Fx + Fy = ma[/tex] (2)
Force in the x axis:
[tex]F_{x} = W_{x}[/tex]
The component of the weight in the x axis can be gotten by means of trigonometric:
[tex]\frac{OC}{H} = sen \theta[/tex]
[tex]\frac{W_{x}}{W} = sen \theta[/tex]
[tex]W_{x} = W sen \theta[/tex]
[tex]W_{x} = mgsen \theta[/tex]
[tex]F_{x} = mgsen \theta[/tex] (3)
Forces in the y axis:
[tex]F_{y} = N - W_{y}[/tex] (4)
The component of the weight in the y axis can be gotten by means of trigonometric:
[tex]\frac{AC}{H} = cos \theta[/tex]
[tex]\frac{W_{y}}{W} = cos \theta[/tex]
[tex]W_{y}= W cos \theta[/tex]
Remember that the weight is defined as:
[tex]W = mg[/tex]
[tex]W_{y}= mgcos \theta[/tex]
The normal force can be obtained from equation (4)
[tex]N - W_{y} = 0[/tex]
[tex]N = W_{y}[/tex]
[tex]N = mgcos \theta[/tex]
Therefore, equation 4 can be rewritten as:
[tex]F_{y} = mgcos \theta - mgcos \theta[/tex]
[tex]F_{y} = 0[/tex] (5)
Then, replacing equation 3 and equation 5 in equation 2 it is gotten:
[tex]mgsen \theta + 0 = ma[/tex]
[tex]mgsen \theta = ma[/tex] (6)
However, a can be isolated from equation 6
[tex] a = \frac{mgsen \theta}{m}[/tex]
[tex] a = gsen \theta[/tex] (7)
Finally, equation 7 can be replaced in equation 2:
[tex]v_{f} = \sqrt{v_{i}^{2} + 2d(gsen \theta)}[/tex]
[tex]v_{f} = \sqrt{(2.6m/s)^{2} + 2(6.0m)(9.8m/s^{2})sen 180^{\circ})}[/tex]
[tex]v_{f} = 2.6m/s[/tex]
Hence, the speed at the bottom of the ramp is 2.6m/s
