Asked by Isabelle
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
PLEASE HELP. I DON'T UNDERSTAND THIS.
please make it detailed
Answers
Answered by
MathMate
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
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
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