Question

What is the minimum speed with which a meteor strikes the top of Earth’s stratosphere (about 41.5 km above the surface), assuming that the meteor begins as a bit of interplanetary debris far from Earth and stationary relative to Earth? Assume that the drag force is negligible until the meteor reaches the stratosphere. Mass of Earth is 5.974 × 1024 kg; radius of Earth is 6.371 × 106 km; and gravitational constant is 6.674 × 10−11 N·m2/kg2.

b. The tension in a ligament in the human knee is approximately proportional to the extension of the ligament, if the extension is not too large.

If a particular ligament has an effective spring constant of 147 N/mm as it is stretched, what is the elastic energy stored in the ligament when stretched by 0.720 cm?

c. A car with mass of 1093 kg accelerates from 0 m/s to 40.0 m/s in 10.0 s. Ignore air resistance. The engine has a 22.0% efficiency, which means that 22.0% of the energy released by the burning gasoline is converted into mechanical energy.

What volume of gasoline is consumed? Assume that the burning of 1.00 L of gasoline releases 46.0 MJ of energy.

d. A particle is constrained to move along the x-axis. The graph describes the potential energy as a function of position.

The particle has a total mechanical energy of −100 J. At time t = 0, the particle is located at x = 8.00 cm and is moving to the left.

What is the particle’s total energy when it is at x = 2.00 cm?

Answer 1

v = minimum speed

r = distance from center of earth = radius of earth + height above the surface = 6.371 x 10⁶ m + 0.415 x 10⁵ m = 6.4125 x 10⁶ m

M = mass of earth = 5.974 x 10²⁴ kg

m = mass of meteor

Using conservation of energy

GMm/r = (0.5) m v²

GM/r = (0.5) v²

(6.67 x 10^-11)(5.974 x 10²⁴)/(6.4125 x 10⁶) = (0.5) v²

v = 11148 m/s

b)

k = spring constant = 147 N/mm = 147 x 10³ N/m

x = stretch in ligament = 0.720 cm = 0.720 x 10^-2 m

Elastic energy stored in ligament is given as

U = (0.5) k x²

U = (0.5) (147 x 10³) ( 0.720 x 10^-2)²

U = 3.81 J