Work-energy theorem word problem?
A 50,000kg space probe is traveling at a speed of 11,000m/s through deep space. The retrorockets are fired along the line of motion to reduce the probe's speed with a force of 400,000N over 2500km. What is the final speed of the probe?
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Work = Change in Kinetic Energy
Fd = 1/2 m vf^2 - 1/2 m vi^2
vf = sqrt(vi^2 + 2Fd/m)
The retrorockets do a negative work since they oppose the space probe movement: W = –F x d = 4 x 10^5 N x 2.5 x 10^6 m = 1.0 x 10^12 J (remember to use an exponential notation with big numbers). This work reduces the kinetic energy of the probe: W = ÎKE = KE(final) – KE(start), where KE(start) = 0.5 x 50,000 kg x sqr (11,000 m/s) = 3.025 x 10^12 J. So KE(final) = W – KE(start) = 2.025 x 10^12 J. In advance, KE(final) = 0.5 x m x sqr (v) and solve for v: v = sqrt (2 x KE(final) / m) = sqrt (2 x 2.025 x 10^12 J / 50,000 kg) = 9,000 m/s = 9 km/s.