Question

A skateboarder is moving 5.25 m/s when he starts to roll up a frictionless hill. How much higher is he when his velocity has reduced to 2.75 m/s. EXPLAIN HELP ME I DON'T GET THIS!!!!

Answer 1

Mechanical energy E = mgh + 1/2mv²

When he starts, let h = 0 ⇒ E₁ = 1/2mv₁²
When he reaches height h ⇒ E₂ = mgh + 1/2mv₂²

Without friction, energy is conserved at all times.

E₁ = E₂
↓
1/2mv₁² = mgh + 1/2mv₂²
↓
1/2v₁² = gh + 1/2v₂²
↓
gh = 1/2(v₁² - v₂²)
↓
h = (v₁² - v₂²) / (2g)

Answer 2

The height is 1.02 m.

Step-by-step explanation:

Given that,

Speed of skateboarder = 5.25 m/s

Reduced velocity = 2.75 m/s

We need to calculate the height

Using mechanical energy,

Without frictional mechanical energy is same all points.

Using conservation of energy

`E_(i)=E_(f)`

`mgh+(1)/(2)mv_(i)^2=mgh+(1)/(2)mv_(f)^2`

When he starts, it means h = 0

So,

`(1)/(2)mv_(i)^2=mgh+(1)/(2)mv_(f)^2`

`(1)/(2)v_(i)^2=gh+(1)/(2)mv_(f)^2`

`h = (v_(i)^2-v_(f)^2)/(2g)`

`h=(5.25^2-2.75^2)/(2*9.8)`

`h=1.02\ m`

Hence, The height is 1.02 m.