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:
a)
b)
Explanation:
Given:
mass of the lighter block,
velocity of the lighter block,
mass of the heavier block,
velocity of the heavier block,
a)
Using conservation of linear momentum:
where:
final velocity of the lighter block
final velocity of the heavier block
........................(1)
Since kinetic energy is conserved in elastic collision:
divide the above equation by eq. (1)
.............................(2)
now we substitute the value of v from eq. (2) in eq. (1)
(negative sign denotes that the direction is towards left)
b)
now we substitute the value of v' from eq. (2) in eq. (1)
Answer:
the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh
Explanation:
Given that;
weight of vehicle = 4000 lbs
we know that 1 kg = 2.20462
so
m = 4000 / 2.20462 = 1814.37 kg
Initial velocity = 60 mph = 26.8224 m/s
Final velocity = 30 mph = 13.4112 m/s
now we determine change in kinetic energy
Δk = m(
² -
² )
we substitute
Δk = ×1814.37( (26.8224)² - (13.4112)² )
Δk = × 1814.37 × 539.5808
Δk = 489500 Joules
we know that; 1 kilowatt hour = 3.6 × 10⁶ Joule
so
Δk = 489500 / 3.6 × 10⁶
Δk = 0.13597 ≈ 0.136 kWh
Therefore, the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh