600.0 mL of air is at 27.0 °C. What is the volume at 60.0 °C?
1 answer:
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a) The speed of the block at a height of 0.25 m is 2.38 m/s
b) The compression of the spring is 0.25 m
c) The final height of the block is 0.54 m
Explanation:
a)
We can solve the problem by using the law of conservation of energy. In fact, the total mechanical energy (sum of kinetic+gravitational potential energy) must be conserved in absence of friction. So we can write:
where
is the initial potential energy, at the top
is the initial kinetic energy, at the top
is the final potential energy, at halfway
is the final kinetic energy, at halfway
The equation can be rewritten as
where:
m = 2.7 kg is the mass of the block
is the acceleration of gravity
is the initial height
u = 0 is the initial speed
is the final height of the block
v is the final speed when the block is at a height of 0.25 m
Solving for v,
b)
The total mechanical energy of the block can be calculated from the initial conditions, and it is
At the bottom of the ramp, the gravitational potential energy has become zero (because the final heigth is zero), and all the energy has been converted into kinetic energy. However, then the block compresses the spring, and the maximum compression of the spring occurs when the block stops: at that moment, all the energy of the block has been converted into elastic potential energy of the spring. So we can write
where
k = 453 N/m is the spring constant
x is the compression of the spring
And solving for x, we find
c)
If there is no friction acting on the block, we can apply again the law of conservation of energy. This time, the initial energy is the elastic potential energy stored in the spring:
while the final energy is the energy at the point of maximum height, where all the energy has been converted into gravitational potetial energy:
where is the maximum height reached. Solving for this quantity, we find
which is the initial height: this is correct, because the total mechanical energy is conserved, so the block must return to its initial position.
Learn more about kinetic and potential energy:
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