Question
An 80 kg crate was slowed to a stop when a frictional force of 160 Newtons was applied. It was originally traveling at 5.5 m/s. What distance did the box travel before it came to a stop? m. Round to first decimal.
Answers
MathMate
v0=5.5 m/s
v1=0 m/s
F=160 N
m=80 kg
Use Newton's second law, F=ma
a=F/m=-2 m/s-2 (deceleration)
Use:
(v1^2-v0^2)=2aS
So
S=(v1^2-v0^2)/2a=30.25/4=7.5625 m
v1=0 m/s
F=160 N
m=80 kg
Use Newton's second law, F=ma
a=F/m=-2 m/s-2 (deceleration)
Use:
(v1^2-v0^2)=2aS
So
S=(v1^2-v0^2)/2a=30.25/4=7.5625 m