Question

A child and sled with a combined mass of 54.8

Answer 1

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, $g = 9.81 m/s^2$.

Angle of the path the horizontal, $\theta = 36$ 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, $F= mg\cdots(i)$

The component of force acting in the direction of motion = $F\sin \theta.$

Let $a$ be the acceleration of the combined mass, m, So,

$F\sin \theta= ma$

$\Rightarrow mg \sin \theta= ma$ [ from equation (i)]

$\Rightarrow a = g \sin \theta \cdots(ii).$

Let v be the final velocity of the combined mass.

Now, by using the equation of motion,

$v^2=u^2+2as\\\\\Rightarrow v^2=0^2+2as\\\\ \Rightarrow v^2=2as\cdots(iii)$

Here, s is the displacement in the direction of motion,

So, s= AB

Now, in the right-angled triangle ABO,

$\sin\theta = OA/AB= h/AB\\\\\Rightarrow AB = h/ \sin\theta\\\\\Rightarrow s = h/ \sin\theta\cdots(iv)$

Now,  from equations (ii), (iii) and (iv), we have

$v^2= 2\times g \sin \theta \times \frac {h}{\sin\theta}\\\\\Rightarrow v^2= 2gh\\\\\Rightarrow v= \sqrt{2gh}$

By using the given values, we have

$v= \sqrt{2\times 9.81\times 11.5}=\sqrt {225.63}\\\\\Rightarrow v = 15.02 m/s$

Hence, the speed of the combined mass at the bottom = 15.02 m/s.