Two fi gure skaters are moving east together during a
performance. Skater 1 (78 kg) is behind skater 2 (56 kg)
when skater 2 pushes on skater 1 with a force of
64 N [W]. Assume that no friction acts on either
skater. T / I
(a) Determine the acceleration of each skater.
(b) What will happen to the motion of each skater?
Explain your reasoning
2 answers
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Fnet = Ft
Skater 1: a = Ft/m
a = 64N/78kg
a = 0.82m/s2[E] kg
Skater 2: a = Ft/m
a = 64N/56kg
a = 1.1 m/s2 [W]
The skaters will accelerate in the opposite direction due to how the action-reaction law states that for every action force there is a simultaneous reaction force equal in magnitude but opposite in direction.
Skater 1: a = Ft/m
a = 64N/78kg
a = 0.82m/s2[E] kg
Skater 2: a = Ft/m
a = 64N/56kg
a = 1.1 m/s2 [W]
The skaters will accelerate in the opposite direction due to how the action-reaction law states that for every action force there is a simultaneous reaction force equal in magnitude but opposite in direction.
4 answers