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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Complete Question:
In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg). Both cars then slide with locked wheels until the frictional force from the slick road (with a low ?k of 0.15) stops them, at distances dA = 6.1 m and dB = 4.4 m. What are the speeds of (a) car A and (b) car B at the start of the sliding, just after the collision? (c) Assuming that linear momentum is conserved during the collision, find the speed of car B just before the collision.
Answer:
a) Speed of car A at the start of sliding = 4.23 m/s
b) speed of car B at the start of sliding = 3.957 m/s
c) Speed of car B before the collision = 7.28 m/s
Explanation:
NB: The figure is not provided but all the parameters needed to solve the question have been given.
Let the frictional force acting on car A, ............(1)
Since frictional force is a type of force, we are safe to say .......(2)
Equating (1) and (2)
a) Speed of A at the start of the sliding
b) Speed of B at the start of the sliding
Let the speed of car B before collision =
Momentum of car B before collision =
Momentum after collision =
Applying the law of conservation of momentum: