What is the best approximate value for the elastic potential energy (EPE) of the spring elongated by 3.0 meters?
1 answer:
The elastic potential energy of the spring is 6.8 J
Explanation:
The elastic potential energy of a compressed/stretched spring is given by the equation:
where
k is the spring constant
x is the elongation of the spring
The spring constant of the spring in this problem can be found by keeping in mind the relationship between force (F) and elongation (x) (Hooke's law):
By looking at the graph and comparing it with the formula, we realize that the slope of the force-elongation graph corresponds to the spring constant. Therefore in this case,
Therefore when the spring has a elongation of , its potential energy is
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Answers:
a) -2.54 m/s
b) -2351.25 J
Explanation:
This problem can be solved by the <u>Conservation of Momentum principle</u>, which establishes that the initial momentum must be equal to the final momentum
:
(1)
Where:
(2)
(3)
is the mass of the first football player
is the velocity of the first football player (to the south)
is the mass of the second football player
is the velocity of the second football player (to the north)
is the final velocity of both football players
With this in mind, let's begin with the answers:
a) Velocity of the players just after the tackle
Substituting (2) and (3) in (1):
(4)
Isolating :
(5)
(6)
(7) The negative sign indicates the direction of the final velocity, to the south
b) Decrease in kinetic energy of the 110kg player
The change in Kinetic energy is defined as:
(8)
Simplifying:
(9)
(10)
Finally:
(10) Where the minus sign indicates the player's kinetic energy has decreased due to the perfectly inelastic collision