A sprinf has a potential energy of 84.08 J and a constant of 342.25 N/m. How far it been stretched? Use potential energy elastic
1 answer:
The spring has been stretched 0.701 m
Explanation:
The elastic potential energy of a spring is the potential energy stored in the spring due to its compression/stretching. It is calculated as
where
k is the spring constant
x is the elongation of the spring with respect to its equilibrium position
For the spring in this problem, we have:
E = 84.08 J (potential energy)
k = 342.25 N/m (spring constant)
Therefore, its elongation is:
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(a) 1.49 m/s
The conservation of momentum states that the total initial momentum is equal to the total final momentum:
where
m = 0.016 kg is the mass of the bullet
is the initial velocity of the bullet
M = 3 kg is the mass of the block
is the initial velocity of the block
v = ? is the final velocity of the block and the bullet
Solving the equation for v, we find
(b) Before: 627.2 J, after: 3.3 J
The initial kinetic energy is (it is just the one of the bullet, since the block is at rest):
The final kinetic energy is the kinetic energy of the bullet+block system after the collision:
(c) The Energy Principle isn't valid for an inelastic collision.
In fact, during an inelastic collision, the total momentum of the system is conserved, while the total kinetic energy is not: this means that part of the kinetic energy of the system is losted in the collision. The principle of conservation of energy, however, is still valid: in fact, the energy has not been simply lost, but it has been converted into other forms of energy (thermal energy).