On the surface of a planet an object is thrown vertically upward with an initial speed of 60 m/s. use a conservation of energy approach to determine the object's speed when the object is at a quarter of it's maximum height. (assume the planet has no atmosphere)
PLEASE HELP. I DON'T UNDERSTAND THIS.
please make it detailed
Start with the conservation of energy (when friction/air resistance are absent).
Kinetic Energy, KE = (1/2)mv²
Potential Energy, PE = mgh
m=mass, v=velocity, h=change in height,
g=acceleration due to gravity.
Conservation of energy states that without external forces and without losses due to dissipative forces such as air resistance and friction, KE+PE=constant at any time.
For the object on an unknown planet, g is unknown (=9.81 m/s² on earth).
On ground,
PE=0, KE=(1/2)mv²
At a quarter of maximum height, one-quarter of its kinetic energy is converted to potential energy, so kinetic energy = (3/4)*(1/2)mv²,
Equate to new velocity v1:
(1/2)mv1²=(3/4)(1/2)mv²
=>
v1²=(3/4)v²
=>
v1=(√3)v/2
To determine the object's speed when it is at a quarter of its maximum height, we can use the principle of conservation of energy. This principle states that the total energy of a system remains constant, assuming no external work or energy losses.
In this case, we can consider the object as a system, and track its potential energy and kinetic energy throughout its motion. The total mechanical energy of the object is given by the sum of its kinetic and potential energy.
At the initial point, when the object is thrown vertically upward, it has only kinetic energy and no potential energy. The kinetic energy of an object is given by the equation:
KE = (1/2)mv^2
where KE represents kinetic energy, m is the mass of the object, and v is the velocity of the object. In this case, we don't have the mass of the object, but we can still proceed using the equation involving velocity alone.
So, at the initial point when the object is thrown vertically upward, the kinetic energy is:
KE_initial = (1/2)m(60 m/s)^2 = 1800m
As the object moves upward, its speed decreases due to the gravitational force acting on it. At its maximum height, the object's vertical velocity becomes 0 m/s. At this point, all of the initial kinetic energy is converted into potential energy, which is given by the equation:
PE = mgh
where PE represents the potential energy, m is the mass of the object, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height of the object above a reference point (in this case, the surface of the planet).
Since we are interested in finding the speed of the object when it is at a quarter of its maximum height, we can further deduce the height at that point.
Let's assume the maximum height reached by the object is h_max. Therefore, a quarter of this height would be:
h_1/4 = (1/4)h_max
Now that we have the height, we can calculate the potential energy at this position:
PE_1/4 = mgh_1/4 = mgh_max/4
From the conservation of energy principle, the total mechanical energy of the system remains constant. Therefore, at the initial point (when KE_initial = 1800m), the total mechanical energy is equal to the sum of potential energy and kinetic energy at the point when the object is at a quarter of its maximum height:
KE_initial = PE_1/4 + KE_1/4
1800m = mgh_max/4 + (1/2)mv_1/4^2
In this equation, we are looking for v_1/4, which represents the speed of the object at a quarter of its maximum height. To solve for v_1/4, we need to rearrange the equation:
1800 = (1/4)gh_max + (1/2)v_1/4^2
Simplifying, we get:
(1/2)v_1/4^2 = 1800 - (1/4)gh_max
v_1/4^2 = (3600 - gh_max/2)
v_1/4 = √(3600 - gh_max/2)
Substituting the value of h_1/4 = (1/4)h_max:
v_1/4 = √(3600 - g(h_max/2))
Since the question assumes the planet has no atmosphere, the acceleration due to gravity (g) remains constant throughout. Therefore, we can substitute this value into the equation to get the final answer.
Please note that in order to determine the value of h_max, you would need additional information, such as the time it takes for the object to reach its maximum height or the maximum height explicitly given.