A car is traveling at 49.0 km/h on a flat highway.
(a) If the coefficient of friction between road and tires on a rainy day is 0.100, what is the minimum distance in which the car will stop?
b) What is the stopping distance when the surface is dry and the coefficient of friction is 0.600?
please give me formulas and numbers!
2006-10-04 15:59:23 · 5 answers · asked by natasha b 1 in Science & Mathematics ➔ Physics
The equation for coefficient of friction (mu), is mu=Ff/Fn, where Ff is the force due to friction and Fn is normal force (mg).
Friction is what slows the car down, so you get Ff=mu * Fn= mu * mg.
Relate this to Newton's second law, which states that Fnet is equal to mass times acceleration. Here's the tricky part. Since the net force is provided by Ff, you get Ff = ma, or mu * mg = ma. Divide by the mass on both sides and you get mu * g = a, or, for the first question, a = .1 * (-9.8 m/s^2)
Now you have the following known quantities:
a= -.98 m/s^2
Vi= 13.61 m/s (convert 49.0 km/h to m/s)
Vf= 0 (because it stops)
To find d, rearrange the equation Vf^2=Vi^2 + 2ad to get d = (Vf^2-Vi^2)/2a. Plug in the known values and you will have your answer.
Part b is the same thing, but a = .6 (-9.8)
Good luck, hope this helps!! :)
2006-10-04 16:00:48 · answer #1 · answered by cushdogjr 3 · 0⤊ 0⤋
Hmm, Something wrong with the 2nd answer.
The mass terms divide out so all you need are the acceleration and velocity terms.
The equation for distance traveled is.
x = (initial velocity)(time) -(1/2)(acceleration)(time^2)
That equation has two unkowns distance and time so you need another equation.
velocity when the car stops is zero the equation for velocity is as follows
velocity = initial velocity - acceleration(time) solve this equation for time
time = velocity initial/ acceleration (in seconds)
velocity = 49 Km/hr or 13.6 m/sec 49x1000/3600
acceleration = 9.8 m/sec^2 (coef of friction)
Plug the time and acceleration numbers into the first equation and crank out the distance for each of the cases.
2006-10-04 16:29:20 · answer #2 · answered by Roadkill 6 · 0⤊ 0⤋
This concern is truly complicated me as well.what's the question? i visit in elementary words guess that the question is: at what attitude is the rope connected to the container. is this the question? if so, answer as follows: because the container strikes at a consistent speed, the consequent or internet rigidity is 0. therefore the utilized rigidity in a horizontal direction should be equivalent to the friction rigidity: utilized rigidity horizontal direction = 25N 25N is the horizontal component to a 43N rigidity at an attitude ? with the floor. sparkling up for ?. cos ? = 25/40 3 cos ? = 0.5814 ? = fifty 4.40 5° above the horizontal The rope develop into at an attitude of fifty 4.40 5° above the horizontal
2016-12-04 07:02:20 · answer #3 · answered by gravitt 4 · 0⤊ 0⤋
a. maximum deceleration without skidding= 1/10 g=1m/s^2, so min. stopping dist.=(13.6^2)/2=92.6 metres. 49kps= 13.6m/s and the formula is s=(v^2)/2a.
b. Using the same formula, stopping distance without skidding= (13.6^2)/12=15.4 metres. But you need to add driver reaction time to both figures.
2006-10-04 16:13:17 · answer #4 · answered by zee_prime 6 · 0⤊ 0⤋
a) the force of friction will produce retardation
f=0.100mg(m=mass of body),a=-0.100mg/m=-0.100g
then put the values in third equation of motion,find the distance
b) put a=-0.600g,calculate using the above procedure
2006-10-04 20:13:35 · answer #5 · answered by vish 1 · 0⤊ 0⤋