conservation of energy example problem and solution.pdf

an example problem about conservation of energy with complete answer solution

1. What will be the speed of a dare devil jumper who is
gliding initially at a height of 500 m at 45 m/s before

diving to a height of 270 m?

Unknown= v;
Given:

» h=500m
+ h=270m
* v=45m/s

Formula:

Conservation of Energy

DERLE; where:

E = Kinetic Energy (K) + Potential Energy (U) where:
K=(1/2)mv2

U=mgh

2. * Derivation of formula

> XE= IE;

> K+ Uys Kyt U; ;

> ;mvi + mgh; = ¿mv +mghy

Our unknown is v; hence we need to transpose mgh, to the other side.
1 2 —-1 2

> zmv; +mgh; - mgh, = mvf

> mo =m Ev? + gh; - ghy) ann Simplify the equation

1 mv? m(3v? + 9h; -ghr)

> Sar mt Divide both sides by m (mass)
> 2% = (Gv? + gh; — ghy) bees Cancel common terms
>2Gv}) =2 G v? + gh; - ghy) ..... Multiply both sides by 2 to

make v; equal to 1

3. * Derivation of formula (cont...)
1 1
D2Gvf) =2 (Gv? + gh; — ghy)
Dvi=vi+2gh;-2gh; nao Combine 2gh
> vf=vi+2g(hi hp) am Get the square root of both sides

> |v?= lv? +2g(h; - hy) to remove the square of v?
> Vs v? +2g(h; — hy) €--- Derived formula to be used.

4. Unknown= v;
Given:

» h=500m
* he=270m
° v=45m/s

. Ya, ¡Ue + 2g(h; — hy)

. | (45m/s)? +2(9.8™ 52) (500m — 270m)

“Vx, (2025""/ 2) +19.6/,2 (500m - 270m)

“Ys (2025/2) + 19.6"/ 2 (230m)

La (2025 "’/ 2) + 45087"/ 2

2
. v=, 6533" / 2
* V;= 80.827 m/s €--- Final speed

5. Given:

* h=500m
» h=270m
° v=45m/s
+ V,=80.827

Check

12 _i 2
¿a + mgh; = mv; + mghs

1

¿vi + gh, = 3% + gh;

0.5(45)? + 9.8(500) = 0.5(80.827) + 9.8 (270)
1012.5 + 4900 = 3266.50 + 2646

+ 5912.5 = 5912.5