Question

A 8.1 kg object initially at rest is pushed down a 15.0 m tall hill. What is the speed of the object at the bottom of the hill?

Answer 1

Answer: 17.14 m/s

Step-by-step explanation:

We can solve this problem applying the Conservation of Mechanical energy (which is the sum of the kinetic energy
`K` and potential energy
`P`), where the total initial mechanical energy
`E_(o)` must be equal to the total final mechanical energy
`E_(f)`:

`E_(o)=E_(f)` (1)

Being:

`E_(o)=K_(o)+P_(o)=(1)/(2)mV_(o)^(2)+mgh_(o)` (2)

`E_(f)=K_(f)+P_(f)=(1)/(2)mV_(f)^(2)+mgh_(f)` (3)

Where:

`m=8.1 kg` is the mass of the object

`V_(o)=0 m/s` is the initial velocity (the object was at rest)

`g=9.8 m/s^(2)` is the acceleration due gravity

`h_(o)=15 m` is the object's initial height or position

`V_(f)` is the final velocity

`h_(f)=0 m` is the object's final height or position

Solving and applying the given conditions:

`(1)/(2)mV_(o)^(2)+mgh_(o)=(1)/(2)mV_(f)^(2)+mgh_(f)` (4)

`mgh_(o)=(1)/(2)mV_(f)^(2)` (5)

Finding
`V_(f)`:

`V_(f)=\sqrt{2gh_(o)}` (6)

`V_(f)=\sqrt{2(9.8 m/s^(2))(15 m)}` (7)

Finally:

`V_(f)=17.14 m/s`