You drop a ball from a height of 2.0 m, and it bounces back to a height of 1.6 m. (a) what fraction of its initial energy is los

Question

Firdavs [7]

3 years ago

10

You drop a ball from a height of 2.0 m, and it bounces back to a height of 1.6 m. (a) what fraction of its initial energy is los

t during the bounce? (b) what is the ball's speed just before and just after the bounce? (c) where did the energy go?

Chemistry

2 answers:

solmaris [256] · Answer 1 · 2022-07-02T22:40:17+03:00

solmaris [256]3 years ago

8 0

Answer:

c to the ball

Explanation:

sweet [91] · Answer 2 · 2022-07-02T21:34:20+03:00

Answer : Part a) Fraction of energy lost : 20 %

Part b) Speed before and after bounce = 6.3 m/s and 5.6 m/s

Part c) Energy is lost as thermal energy .

Part A) Fraction of energy lost during bouncing :

The energy possessed by any object when present at any height is potential energy . The formula of potential energy is given as :

PE = mgh

where PE = potential energy

,m = mass pf object , g = gravitational acceleration and h = height

Given : Initial height , h₁ = 2 m final height , h₂ = 1.6 m

Initial potential energy : m * g* h ₁

Final potential energy = m* g* h₂

Energy lost = Initial PE - Final PE

= ( mgh₁ - mgh2 )

Fraction of energy lost : $\frac{energy lost}{initial energy}=\frac{mgh1 - mgh2}{mgh1}$

Plugging value in above formula and taking " mg " common =>

Fraction of energy lost = $\frac{mg(h1-h2)}{mg (h1)} * 100$

$= \frac{(h1-h2)}{h1} * 100$

$= \frac{(2 - 1.6) }{2} * 100$

Fraction of energy lost = 20%

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Part B ) Speed of ball just before and after the bounce.

Speed of ball before the bounce :

The potential energy gets converted to kinetic energy when it fall from height of 2m , so

Potential energy = kinetic energy

mgh₁ = $\frac{1}{2}$ m v²

or v ² = 2gh₁

Given : g = 9.8 m/s² h= 2 m

v² = 2 * 9.8 m/s² * 2 m = 39.2 m²/s²

v = 6.3 m/s

Speed of ball after bounce :

Potential energy = kinetic energy

mgh₂ = $\frac{1}{2}$ m v²

or v² = 2gh₂

= 2 * 9.8 m/s² * 1.6 m = 31.36 m²/s²

v = 5.6 m/s

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Part C) The energy lost due to friction. When the ball touches the ground , there occur friction force between the surface of ground and ball , due to which energy is lost as thermal energy .