A) Elastic potential energy: 1 J
B) Kinetic energy: 1 J
C) Velocity of the block: 1.63 m/s
D) Energy lost due to friction: 0.35 J
Explanation:
A)
The elastic potential energy of a spring is given by the equation
[tex]E=\frac{1}{2}kx^2[/tex]
where
k is the spring constant
x is the compression of the spring
In this problem, we have:
k = 50 N/m is the spring constant
x = 0.2 m is the compression of the spring
Therefore, the elastic potential energy is
[tex]E=\frac{1}{2}(50)(0.2)^2=1 J[/tex]
B)
In this case, the surface is frictionless.
This means that there is no friction force acting on the wood block. Therefore, this also means that the total mechanical energy of the system, which is the sum of kinetic and potential energy, is constant:
[tex]E=PE+KE[/tex]
where
PE is the elastic potential energy
KE is the kinetic energy
When the spring is released, the spring returns to its original length, therefore x = 0; so, all the initial potential energy is converted into kinetic energy, therefore the kinetic energy of the block is
KE = 1 J
c)
The kinetic energy of an object is the energy possessed by the object due to its motion, and it is given by
[tex]KE=\frac{1}{2}mv^2[/tex]
where
m is the mass of the object
v is its velocity
For the block in this problem,
m = 0.75 kg
Therefore, since the kinetic energy is
KE = 1 J
The velocity of the block is
[tex]v=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(1)}{0.75}}=1.63 m/s[/tex]
D)
If the block leaves the spring with a velocity of
v = 1.32 m/s
Then it means that its kinetic energy is
[tex]KE'=\frac{1}{2}mv^2=\frac{1}{2}(0.75)(1.32)^2=0.65 J[/tex]
This also means that the total mechanical energy of the system as the block is released is
E' = 0.65 J
We said instead that the total mechanical energy when the spring was compressed was
E = 1 J
Therefore, the energy lost due to friction is
[tex]\Delta E=E-E'=1-0.65=0.35 J[/tex]
Learn more about kinetic and potential energy:
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