3.A horizontal spring-mass system has low friction, spring stiffness 175 N/m, an

Question

3.A horizontal spring-mass system has low friction, spring stiffness 175 N/m, and mass 0.4 kg. The system is released with an initial compression of the spring of 12 cm and an initial speed of the mass of 3 m/s.

(a) What is the maximum stretch during the motion?
(b) What is the maximum speed during the motion?
(c) Now suppose that there is energy dissipation of 0.02 J per cycle of the spring-mass system. What is the average power input in watts required to maintain a steady oscillation?

4. A spacecraft is coasting toward Mars. The mass of Mars is 6.4 × 1023 kg and its radius is 3400 km (3.4 × 106 m). When the spacecraft is 8700 km (8.7 × 106 m) from the center of Mars, the spacecraft's speed is 2500 m/s. Later, when the spacecraft is 4900 km (4.9 × 106 m) from the center of Mars, what is its speed? Assume that the effects of Mars's two tiny moons, the other planets, and the Sun are negligible. Precision is required to land on Mars, so make an accurate calculation, not a rough, approximate calculation.

Explanation / Answer

1. when on the surface of asteroid,

Potential energy, PEi= - G M m / d

   = - (6.67 x 10^-11 x 10 x 8 x 10^20 ) / (1.9 x 10^5)

    = - 2808421.05 J

KE = m v^2 /2 = 5v^2

when very far away,

PEf = 0

KE = 10 x 169^2 /2 = 142805 J

Using energy conservation,

PEi + KEi = Pef + KEf

- 2808421.05 + 5v^2 = 0 + 142805

v = 768.27 m/s

2. initially PE of system = 0

initial KE = 2 x 0.19 MeV = 0.38 x 10^6 x 1.6 x 10^-19 J = 6.08 x 10^-14 J

finally when they stops,

KEf = 0

PEf = kq1q2/d

= ( 9 x 10^9 x 1.6 x 10^-19 x 1.6 x 10^-19 ) / d

= 2.304 x 10^-28 / d

Using energy conservation,

0 + 6.08 x 10^-14 = 2.304 x 10^-28 / d + 0

d = 3.789 x 10^-15 J

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