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1. What is the potential energy of a 4-kilogram potted plant that is on a 1 meter-high plant
1 answer:
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(a) 2446 N/m
When the spring is at its maximum displacement, the elastic potential energy of the system is equal to the total mechanical energy:
where
U is the elastic potential energy
k is the spring constant
A is the maximum displacement (the amplitude)
Here we have
U = E = 50.9 J
A = 0.204 m
Substituting and solving the formula for k,
(b) 50.9 J
The total mechanical energy of the system at any time during the motion is given by:
E = K + U
where
K is the kinetic energy
U is the elastic potential energy
We know that the total mechanical energy is constant: E = 50.9 J
We also know that at the equilibrium point, the elastic potential energy is zero:
because x (the displacement) is zero
Therefore the kinetic energy at the equilibrium point is simply equal to the total mechanical energy:
(c) 8.55 kg
The maximum speed of the block is v = 3.45 m/s, and it occurs when the kinetic energy is maximum, so when
K = 50.9 J (at the equilibrium position)
Kinetic energy can be written as
where m is the mass
Solving the equation for m, we find the mass:
(d) 2.14 m/s
When the displacement is
x = 0.160 m
The elastic potential energy is
So the kinetic energy is
And so we can find the speed through the formula of the kinetic energy:
(e) 19.6 J
The elastic potential energy when the displacement is x = 0.160 m is given by
And since the total mechanical energy E is constant:
E = 50.9 J
the kinetic energy of the block at this point is
(f) 31.3 J
The elastic potential energy stored in the spring at any time is
where
k = 2446 N/m is the spring constant
x is the displacement
Substituting
x = 0.160 m
we find the elastic potential energy:
(g) x = 0
The postion at that instant is x = 0, since it is given that at that instant the system passes the equilibrium position, which is zero.