A cat with a mass of 4.50 kilograms sits on a ledge 0.800 meters above the ground. If it jumps to the ground, how much kinetic e
2 answers:
The correct answer to the question is A) 35.3 J i.e the kinetic energy of the cat is 35.3 J.
CALCULATION:
As per the question, the mass of the cat m = 4.50 kg.
The cat sits on a ledge which is 0.800 m above the ground.
Hence, the height of the cat from the surface of earth h = 0.800 m.
We are asked to determine the kinetic energy of the cat when it just reaches the ground.
The potential energy of the cat at a height h is calculated as -
Potential energy P.E = mgh
= 4.50 × 9.8 × 0.800 J.
= 35.28 J
= 35.3 J
Here, g is called acceleration due to gravity. Its value is g =
From law of conservation of energy, we know that energy is neither created nor destroyed. The total energy of a system is always constant, and energy changes from one form to another form.
When the cat is at certain height from the ground, it has potential energy. When it starts jumping, its potential energy is converted into kinetic energy. When it just reaches the ground, whole of its potential energy must have converted into kinetic energy.
Hence, the kinetic energy of the cat just before reaching ground = Potential energy of the cat at height 0.800 m above the ground.
Hence, the kinetic energy of the cat just before reaching the ground is 35.3 Joule.
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Answer:
The effective spring constant of the firing mechanism is 1808N/m.
Explanation:
First, we can use kinematics to obtain the initial velocity of the performer. Since we know the angle at which he was launched, the horizontal distance and the time in which it's traveled, we can calculate the speed by:
(This is correct because the horizontal motion has acceleration zero). Then:
Now, we can use energy to obtain the spring constant of the firing mechanism. By the conservation of mechanical energy, considering the instant in which the elastic band is at its maximum stretch as t=0, and the instant in which the performer flies free of the bands as final time, we have:
Then, plugging in the given values, we obtain:
Finally, the effective spring constant of the firing mechanism is 1808N/m.