Answer:
15.02 m/s.
Explanation:
Given that the height of the hill, h= 11.5 m.
Combined mass, m= 54.8 kg
The initial velocity of the combined mass, u=0
Acceleration due to gravity, [tex]g = 9.81 m/s^2[/tex].
Angle of the path the horizontal, [tex]\theta = 36[/tex] degree.
Let A be the initial position and B be the final position of the sled as shown in the figure.
The path is frictionless so the drag force =0
The gravitational force acting on the combined mass in the downward direction, [tex]F= mg\cdots(i)[/tex]
The component of force acting in the direction of motion = [tex]F\sin \theta.[/tex]
Let [tex]a[/tex] be the acceleration of the combined mass, m, So,
[tex]F\sin \theta= ma[/tex]
[tex]\Rightarrow mg \sin \theta= ma[/tex] [ from equation (i)]
[tex]\Rightarrow a = g \sin \theta \cdots(ii).[/tex]
Let v be the final velocity of the combined mass.
Now, by using the equation of motion,
[tex]v^2=u^2+2as\\\\\Rightarrow v^2=0^2+2as\\\\ \Rightarrow v^2=2as\cdots(iii)[/tex]
Here, s is the displacement in the direction of motion,
So, s= AB
Now, in the right-angled triangle ABO,
[tex]\sin\theta = OA/AB= h/AB\\\\\Rightarrow AB = h/ \sin\theta\\\\\Rightarrow s = h/ \sin\theta\cdots(iv)[/tex]
Now, from equations (ii), (iii) and (iv), we have
[tex]v^2= 2\times g \sin \theta \times \frac {h}{\sin\theta}\\\\\Rightarrow v^2= 2gh\\\\\Rightarrow v= \sqrt{2gh}[/tex]
By using the given values, we have
[tex]v= \sqrt{2\times 9.81\times 11.5}=\sqrt {225.63}\\\\\Rightarrow v = 15.02 m/s[/tex]
Hence, the speed of the combined mass at the bottom = 15.02 m/s.
