F = m a
if velocity is constant, a = 0
so
F = 0
Therefore
the force is exactly equal and opposite
2.
a = change in velocity/change in time
a = (45-20)/.7 = 35.71
F = m a = .5*35.71 = 17.85 N so yes
3. change in velocity = 20-50 = -30 km/hr
-30 km/hr * (1000 m/km)(1 hr/3600s)
= -8.33 m/s
so
a = -8.33/time
to find time
average speed = (50+20)/2 = 35 km/hr
35*(1000/3600) = 9.72 m/s
so
time deaccelerating = 30m/9.72m/s
= 3.09 seconds
so acceleration = -8.33/3.09 = -2.7 m/s^2
F = m a = 800 (-2.7) = -2159 Newtons
NOTE - this is not really right in the real world of boats. The boat is in water some of which partners in the acceleration of the boat creating what naval architects call "added mass" and a larger force would actually be required.
1. Harry is pushing a car down a level road at 2.0m/s with a force of 243N. The total force acting on the car in the opposite direction, including road friction and air resistance, is which of the following?
a) slightly more than 243N
b) exactly equal to 243N
c) slightly less than 243N
2. a 0.50kg model rocket accelerates from 20m/s [up] to 45 m/s [up] in 0.70s. Calculate the unbalanced force acting on it. (I got 18N. Is that correct?)
3. an 800km boat slows down uniformly from 50km/h [E] to 20km/h [E] as it enters the harbour. If the boat slows down over a 30m distance, what is the force of friction on the boat? (hint, convert velocities to m/s)
1 answer