A 66.0 kg boy and a 47.0 kg girl use an elastic rope while engaged ina tug-of-war on an icy, frictionless surface. If the acceleration of the girl towards the boy is 2.31 m/s^2, what is the magnitude of the acceleration of the boy towards the girl?
2006-09-15 08:08:35 · 7 answers · asked by beautyqueenjustine 3 in Science & Mathematics ➔ Physics
If the girl is given acceleration towards the boy there must be tension in the rope. The force is
F = ma = 47 kg*2.31 m/s^2.
The boy would have the same force on him so
F = 47 kg*2.31 m/s^2 = 66.0 kg*a
a = (47 kg*2.31 m/s^2) / 66.0 kg = 1.645 m/s^2
2006-09-15 08:40:45 · answer #1 · answered by sojsail 7 · 0⤊ 0⤋
same 2.31 m/s^2 ?? if i imagine 2 bodies accelerating towards each other by gravity , they approch each other at a specific acc.
anyway , once the girl starts moving she cannot get purchase on the ground for any more tug of war. ( it is an icy frictionless surface )
2006-09-15 15:18:20 · answer #2 · answered by jaco 3 · 0⤊ 0⤋
Here's the story. When 2 people pull in opp directions , the force in both directions is equal. Since the a of the girl is 2.31m/s/s and her m=47 kg, we can find her f by usinf F=MA. F=(47kg)(2.31 m/s/s) which equals 108.57n
Since this is also the force on the boy and his M=66kg, we can find his A by using F=MA again. So, 108.57n=(66) (A). Solve for A and you have it...........
2006-09-15 16:23:48 · answer #3 · answered by jay k 1 · 0⤊ 0⤋
Since it is elastic rope I believe the acceleration is zero since the greater force will make girl go towards boy
2006-09-15 15:43:12 · answer #4 · answered by Anonymous · 0⤊ 0⤋
they should stop playing tug of war and jump into bed together
2006-09-15 15:10:45 · answer #5 · answered by jessop m 2 · 0⤊ 0⤋
is there a "Caution Slippery When Wet" sign up?
2006-09-15 15:16:13 · answer #6 · answered by D G M 1 · 0⤊ 0⤋
zero
2006-09-15 15:13:01 · answer #7 · answered by clyde 3 · 0⤊ 0⤋