Question

If a 3-kg rabbit's leg muscles act as imperfectly elastic springs, how much energy will they hold if the rabbit lands from a height of 0.5 m and its legs are compressed by 0.2 m?

Answer 1

U = 14.7 J

Step-by-step explanation:

Loss in gravitational potential energy is stored in the form of potential energy of rabbit legs

So here we will have by energy conservation

gravitational potential energy = energy stored in the legs

so here we will have

gravitational potential energy given as

`U = mgh`

m = 3 kg

g = 9.81 m/s/s

h = 0.5 m

now we have

`U = 3(9.81)(0.5)`

`U = 14.7 J`

Answer 2

Answer;

- 15 J

Explanation;

-Potential energy is defined as mechanical energy, stored energy, or energy caused by its position.

-For the gravitational force the formula is P.E. = mgh, where m is the mass in kilograms, g is the acceleration due to gravity (9.8 m /s² at the surface of the earth) and h is the height in meters.

Potential energy of the rabbit at the peak of its height is

PE = (3)(10)(0.5) = 15 J

(around 14.7 but because energy is lost, it is less than that)