Answer:
The compression of the spring is 0.633 m.
Explanation:
Given;
mass of the skier, m = 70 kg
speed of the skier, v = 4 m/s
spring constant of the spring, k = 2800 N/m
At the bottom of the hill the kinetic energy of the skier will be maximum while the potential energy will be zero, thus all the mechanical energy of the skier will be converted to kinetic energy.
Apply the law of conservation of energy;
the kinetic energy of the skier at the bottom hill = elastic potential energy of the spring.
[tex]\frac{1}{2}mv^2 = \frac{1}{2}kx^2\\\\mv^2 = kx^2\\\\x^2 = \frac{mv^2}{k} \\\\x = \sqrt{\frac{mv^2}{k}}[/tex]
where;
x is compression of the spring
[tex]x = \sqrt{\frac{mv^2}{k}}\\\\x = \sqrt{\frac{(70)(4)^2}{2800}}\\\\x = 0.633 \ m[/tex]
Therefore, the compression of the spring is 0.633 m.
