Question

A child and sled with a combined mass of 49.0 kg slide down a frictionless hill that is 7.50 m high at an angle of 26 ◦ from horizontal. The acceleration of gravity is 9.81 m/s 2 . If the sled starts from rest, what is its speed at the bottom of the hill?

Answer 1

Answer:12.12 m/s

Explanation:

Given

mass of child $m=49 kg$

height of hill $h=7.5 m$

inclination $\theta =26^{\circ}$

Conserving Energy at top and bottom Point of hill

Potential Energy at Top =Kinetic Energy at bottom

$mgh=\frac{mv^2}{2}$

$v=\sqrt{2gh}$

$v=\sqrt{2\times 9.8\times 7.5}$

$v=\sqrt{147}$

$v=12.12 m/s$

Answer 2

Answer:

$v\approx 12.129\,\frac{m}{s}$

Explanation:

The final speed of the child-sled system is determined by means of the Principle of Energy Conservation:

$U_{1} + K_{1} = U_{2} + K_{2}$

$K_{2} = (U_{1}-U_{2})+K_{1}$

$\frac{1}{2}\cdot m \cdot v^{2} = m\cdot g \cdot \Delta h$

$v = \sqrt{2\cdot g \cdot \Delta h}$

$v = \sqrt{2\cdot\left(9.807\,\frac{m}{s^{2}} \right)\cdot (7.50\,m)}$

$v\approx 12.129\,\frac{m}{s}$