A crate with a mass of 110 kg glides through a space station with a speed of 2 m/s. An astronaut speeds it up by pushing on it from behind with a force of 240 N, continually pushing with this force through a distance of 4 m. The astronaut moves around to the front of the crate and slows the crate down by pushing backwards with a force of 230 N, backing up through a distance of 3 m. After these two maneuvers, what is the speed of the crate?
final speed =_________ m/s
To find the final speed of the crate, we need to apply the concept of work and energy. The work done on an object is equal to the change in its kinetic energy.
1. Let's calculate the work done by the astronaut while pushing the crate forward:
Work done = force x distance
Work done = 240 N x 4 m
Work done = 960 Joules
2. The work done by the astronaut increases the kinetic energy of the crate. So, the increase in kinetic energy is equal to the work done:
ΔKE = 960 J
3. We can use the equation for kinetic energy to calculate the initial kinetic energy of the crate:
KE = 0.5 x mass x velocity^2
0.5 x 110 kg x (2 m/s)^2 = 220 Joules
4. The final kinetic energy of the crate will be the sum of the initial kinetic energy and the increase in kinetic energy:
Final KE = Initial KE + ΔKE
Final KE = 220 J + 960 J
Final KE = 1180 Joules
5. To find the final speed, we need to rearrange the kinetic energy equation:
Final KE = 0.5 x mass x final velocity^2
1180 J = 0.5 x 110 kg x final velocity^2
Divide both sides by (0.5 x 110 kg):
final velocity^2 = 1180 J / (0.5 x 110 kg)
final velocity^2 = 21.45 m^2/s^2
6. Finally, take the square root of both sides to find the final velocity:
final velocity = √(21.45 m^2/s^2) ≈ 4.63 m/s
Therefore, the final speed of the crate is approximately 4.63 m/s.