Question

B) A 1300 -kg car is pushing an out-of-gear 2200 -kgtruck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push horizontally against the ground with a force of 4500 N . The rolling friction of the car can be neglected, but the heavier truck has a rolling friction of 760 N , including the "friction" of turning the truck's drivetrain. What is the acceleration aT of the truck?

C) An 89.0 kg spacewalking astronaut pushes off a 620 kg satellite, exerting a 85.0 N force for the 0.540 s it takes him to straighten his arms. How far apart are the astronaut and the satellite after 1.50 min ?

Answer 1

(B)by newton law

Fnet=(1300+2200)*a

4500-760=3500a

a=1.068m/s^2

(C)

The relative velocity of the satellite to the astronaut is zero. So the initial velocity of the satellite relative to the astronaut after his push by the force he applied is zero.  

Now F = 639*a
a = F/m = 85/620= 0.137 m/s^2. This acceleration lasts only that same timing during which that astronaut applied his force on the satellite.

V final = v initial +at = v final = 0.137 *0.540 seconds= 0.074 m/s
After 1.5 minutes which is converted to 90 seconds, the satellite will travel 0.074*90 = 6.66 meters

Since to each action there is a reaction the same force applied on the satellite will be applied on the astronaut and will cause him to move backwards in accordance with the following equation:

F = m*a. a = F/m = 85/89 =0.9550m/s^2
V final = 0.9550*0.540= 0.5157 m/s
After 1.5 min the astronaut will have backed away 46.413 meters
The total separation between the astronaut and the satellite will be 46.413 meters + 6.66 meters
Separation = 53.073 meters away