Two ice skaters of weight 60 kg and 50 kg are holding the two ends of a rope.The rope is taut. The 60 kg man pulls the rope with 20 N force. What will be the force exerted by the
rope on the other person? What will be their respective acceleration?
Answers
Answered by
4
The tension is propogated and hence the force across the string remains as 20N which is also the force applied on the second man
The respective acceleration will be
1: 20/60 = 1/3
2: 20/50 = 2/5
Answered by
6
1.mass = 60 kg,F = 20 N
F=m×a
20=60×a
a=20/60
a=2/6=0.3 m/s²
2.mass = 50kg,F = 20 N
F=m×a
20=50×a
a=20/50
a=2/5=0.4 m/s²
F=m×a
20=60×a
a=20/60
a=2/6=0.3 m/s²
2.mass = 50kg,F = 20 N
F=m×a
20=50×a
a=20/50
a=2/5=0.4 m/s²
Similar questions