Is it possible for a system to have negative potentialenergy?
1 answer:
Answer:
Yes, since the choice of the zero o potential energy is arbitrary.
Explanation:
The kinetic energy is due to the motion of the object. The expression for the kinetic energy is as follows;
Here, m is the mass of the object and v is the velocity of the object.
The kinetic energy can not be negative as the velocity is squared. It can be zero and positive.
Potential energy: It is the energy is due to the position of the object.
The expression for the potential energy is as follows;
PE= mgh
Here, g is the acceleration due to gravity and height.
Height can be taken from the reference point, zero which can be taken below zero and above zero. Zero is taken as origin. Below zero, the height is taken as negative and above zero, the height is taken as positive.
The potential energy can be zero, positive and negative.
The total energy is the sum of kinetic energy and potential energy.
E= KE + PE
Here, KE is the kinetic energy and PE is the potential energy.
Therefore, the option (B) is correct.
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The rocket travelled a maximum height at 1.0102 km.
Given,
The acceleration of a rocket (a) = 12 m/s²
The altitude of the rocket (s) = 0.50 km = 0.5×10³m
The maximum height of the rocket (h) = ?
Solution,
A rocket is a spacecraft, aircraft, vehicle or projectile that obtains thrust from a rocket engine.
The rate of change of the velocity of an object with respect to time is known as acceleration. It is denoted by (a).i.e. unit is m/s²
(a) = Δv/Δt
Where , Δv is change in velocity and Δt is change in time.
The rate of change in position with respect to time is known as velocity. i.e. Its unit is m/s.
(v)= Δx/Δt
Where,Δx is the change in position and Δt is change in time & v is velocity.
Therefore we know the equation of motion is written as,
v² = u² +2as
Where, v is final velocity , u is initial velocity , a is acceleration and s is altitude of the rocket.
Then putting the value ,
v² = 0 + ( 2× 10 × 0.5×10³)m/s
v² = m/s
v = 100 m/s
Therefore, at altitude of 0.50 km the initial velocity of rocket (u) will be 100 m/s, final velocity v become zero and under free falling the acceleration will be taken (-g) then equation of motion can be given as ,
v² = u² - 2(g)h
h = (v²- u² ) / 2g
h = 10,000/2×9.8
h = 510.2 m
So that the rocket travelled the maximum height ,
(h)= (0.5 km + 510.2m)
(h) = 1.0102 km
Hence, the rocket travelled at the maximum height h is 1.0102 km
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