Answer:
a) 0.0625
b) 0.177 m/s
Explanation:
The spring has potential energy stored, when it pushes on the ice block it will use that energy to perform work. The work will be converted into kinetic energy on the ice block.
The elastic potential energy on the spring is:
Ep = 1/2 * k * Δx^2
Ep = 1/2 * 200 * 0.025^2 = 0.0625 J
That is the energy that will be transferred as kinetic energy on the ice block.
Ec = 1/2 * m * v^2
Rearranging:
v^2 = 2 * Ec / m
[tex]v = \sqrt{ 2 * Ec / m}[/tex]
[tex]v = \sqrt{ 2 *0.0625 / 4} = 0.177 m/s[/tex]
A. The work done on the block by the spring is 0.0625 J
B. The speed of the block after it leaves the spring is 0.18 m/s
The workdone by the spring can be obtained as follow:
Wd = ½Ke²
Wd = ½ × 200 × 0.025²
Wd = 100 × 0.000625
Wd = 0.0625 J
E = ½mv²
0.0625 = ½ × 4 × v²
0.0625 = 2 × v²
Divide both side by 2
v² = 0.0625 / 2
v² = 0.03125
Take the square root of both side
v = √0.03125
v = 0.18 m/s
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