In an Olympic fihure skating event, a 65 kg male skater pushes a 45 kg female skater, causing her to accelerate at a rate of 2.0 m/s^2. At what rate will the male skater accelerate ? What is the direction of his acceleration? I know this questions deals with Newtons laws but I cannot figure out what to do to find the answer? Please help.
2007-03-21 13:29:34 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Whenever male skater exerts a force on female skater , female skater simultaneously exerts a force on him with the same magnitude in the opposite direction. (Third law of motion)
F1=F2
m1a1=m2a2
a1= (m2/m1)a2
a2= (45/65) 2=1.38m/s^2 in the apposite direction then acceleration of 2 m/s^2
2007-03-21 13:42:00 · answer #1 · answered by Edward 7 · 0⤊ 0⤋
the force that the male exerts on the female is equal to the force the female exerts on the male (newton's 3rd law)
Fmale = Ffemale
ma1 = ma 2
(65kg)a =(45kg)(2m/s^2)
(65kg)a = 90N
a = 1.385m/s^2.
the direction of the male's acceleration is in the opposite direction of the female's acceleration.
hope this helps
2007-03-21 20:44:05 · answer #2 · answered by 7 · 0⤊ 0⤋
Hmmm... well acceleration= final velocity- original velocity i believe? that might help...i did that last year. i dont really remember a lot of it now...sorry!
2007-03-21 20:41:54 · answer #3 · answered by bluesclues 2 · 0⤊ 0⤋