F = m Acc
mu m g = m v^2/R
v^2 = mu g R
A car moves on a level horizontal road in
a circle of radius 30.5 m. the coefficient of
friction between the tires and the road is 0.5.
The maximum speed with which the car
can round this curve without slipping is:
6 answers
its 67
3 m/s
its clearly 13m/s
v = sqrt( mu * g * r)
v = sqrt( .5 * 9.8 * 30.5)
v = 12.2m/s
v = sqrt( .5 * 9.8 * 30.5)
v = 12.2m/s
The correct answer is 12.2 m/s
How?
To do it you need the equation μ x m x g = (m x v^2)/R
Since we are trying to find the velocity or the maximum speed of the car we would change the equation to v^2 = μ g R
μ = 0.5 which is the friction between the tires and the road
g = gravitational acceleration (9.8 m/s^2)
r = is the radius which is 30.5 m
Once plugged into the equation it should look like this:
v^2 = (0.5 x 9.8 x 30.5)
so v^2 = 149.45
you need to square root both sides so it would be v = √(149.45)
so the answer is 12.22497444 m/s but rounded it would be 12.22 m/s
How?
To do it you need the equation μ x m x g = (m x v^2)/R
Since we are trying to find the velocity or the maximum speed of the car we would change the equation to v^2 = μ g R
μ = 0.5 which is the friction between the tires and the road
g = gravitational acceleration (9.8 m/s^2)
r = is the radius which is 30.5 m
Once plugged into the equation it should look like this:
v^2 = (0.5 x 9.8 x 30.5)
so v^2 = 149.45
you need to square root both sides so it would be v = √(149.45)
so the answer is 12.22497444 m/s but rounded it would be 12.22 m/s
0 answers