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Andrei [34K]
3 years ago
8

A young boy of mass m = 25 kg sits on a coiled spring that has been compressed to a length 0.4 m shorter than its uncompressed l

ength and then held at this length. Suddenly the spring is released, and the boy flies vertically into the air. He reaches a maximum distance 0.5 m above his initial position. The spring is ideal and massless and we ignore the air friction.
What is the spring constant k of the spring?

What is the speed of the boy when he is 0.4 meters above his starting position?
Physics
1 answer:
rewona [7]3 years ago
3 0
The spring's elastic potential energy is converted into elastic potential energy of the boy.
Potential energy = elastic energy
mgh = 1/2 kx²
k = (2 * 25 * 9.81 * 0.5) / 0.4²
k = 1533 N/m

We will apply
2as = v² - u², with u = 0
a = F/m
F = kx
a = kx/m

v = √(2 x 0.4 x (0.4 x 1533 / 25))
v = 4.43 m/s
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v_A=7667m/s\\\\v_B=7487m/s

Explanation:

The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

F_g=\frac{GMm}{R^{2} }

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.

Also, we know that the centripetal force of an object describing a circular motion is given by:

F_c=m\frac{v^{2}}{R}

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So R_A=6774km=6.774*10^6m and R_B=7103km=7.103*10^6m (Since R_{earth}=6371km). Then, we get:

v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).

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