Question
A curve in a road forms part of a horizontal circle. As a car goes around it at constant speed 14.0 m/s, the total force exerted on the driver has magnitude 115 N. What are the magnitude and direction of the total vector force exerted on the driver if the speed is 25.0 m/s instead?
I thought you would use proportions, but its not right. THANK YOU
I thought you would use proportions, but its not right. THANK YOU
Answers
Damon
Ignoring gravity
v^2/R = centripetal acceleration
So force toward center = mv^2/R
115 N = m 14^2/R
so
m/R = 115/14^2
F = m 25^2/R = (115/14^2) (25^2)
v^2/R = centripetal acceleration
So force toward center = mv^2/R
115 N = m 14^2/R
so
m/R = 115/14^2
F = m 25^2/R = (115/14^2) (25^2)
Laila
Thank you so much! Makes perfect sense =D
Damon
You are welcome :)
aoun raza
In ∑F=m
r
v
2
, both m and r are unknown but remain constant.
Symbolically, write
∑F
slow
=(
r
m
)(14.0m/s)
2
and ∑F
fast
=(
r
m
)(18.0m/s)
2
Therefore, ∑F is proportional to v
2
and increases by a factor of (
14.0
18.0
)
2
as v increases from 14.0m/s to 18.0m/s. The total force at the higher speed is then
∑F
fast
=(
14.0
18.0
)
2
∑F
slow
=(
14.0
18.0
)
2
(130N)=215N
r
v
2
, both m and r are unknown but remain constant.
Symbolically, write
∑F
slow
=(
r
m
)(14.0m/s)
2
and ∑F
fast
=(
r
m
)(18.0m/s)
2
Therefore, ∑F is proportional to v
2
and increases by a factor of (
14.0
18.0
)
2
as v increases from 14.0m/s to 18.0m/s. The total force at the higher speed is then
∑F
fast
=(
14.0
18.0
)
2
∑F
slow
=(
14.0
18.0
)
2
(130N)=215N