Respuesta :
Answer:
The velocity at the bottom of the ramp is 7.32 m/s
Solution:
As per the question:
Mass of the object, m = 200 kg
Length of the rope, l = 8 m
Angle of inclination of the ramp, [tex]\theta = 20^{\circ}[/tex]
g = 9.81 [tex]m/s^{2}[/tex]
Now,
The overall mechanical energy of the system will remain conserved:
Mechanical energy at the top = Mechanical energy at the bottom
[tex]KE_{top} + PE_{top} = KE_{bot} + PE_{bot}[/tex] (1)
where
[tex]KE_{top}[/tex] = Kinetic Energy at the top
[tex]PE_{top}[/tex] = Potential Energy at the top
[tex]KE_{bot}[/tex] = Kinetic Energy at the bottom
[tex]KE_{bot}[/tex] = Kinetic Energy at the bottom
Potential energy at the top is min and potential energy at the top is maximum whereas potential energy at the bottom is minimum and kinetic energy is maximum
Thus, using eqn (1):
[tex]\frac{1}{2}mv^{2}_{top} + mgh_{top} = \frac{1}{2}mv^{2}{bot} + mgh_{bot}[/tex]
where velocity at the top is minimum and zero and height at the bottom is zero.
Also, the object slides along the inclined plane and hence the height, h = [tex]l\sin\theta[/tex]
Thus
[tex]\frac{1}{2}m.0^{2}_{top} + 200\times 9.8\times lsin\theta = \frac{1}{2}mv^{2}{bot} + mg.0[/tex]
[tex]200\times 9.8\times 8sin20 = \frac{1}{2}\times 200v^{2}{bot}[/tex]
[tex]v_{bot} = \sqrt{2\times 26.81} = 7.32\ m/s[/tex]
