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

`v\approx 12.129\,(m)/(s)`

Step-by-step 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)`

`(1)/(2)\cdot m \cdot v^(2) = m\cdot g \cdot \Delta h`

`v = √(2\cdot g \cdot \Delta h)`

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

`v\approx 12.129\,(m)/(s)`

Answer 2

Answer:12.12 m/s

Step-by-step 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=(mv^2)/(2)`

`v=√(2gh)`

`v=√(2* 9.8* 7.5)`

`v=√(147)`

`v=12.12 m/s`