Physics help

[byeh2004]

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A car is traveling at 51.0mi/h on a horizontal highway

If the coefficient of static friction between the road and tires on a rainy day is 0.097, what is the minimum distance in which the car will stop?

what is the stopping distance when the surface is dry and the coffefficent of static friction = 0.605?

 

[mercaptovizadeh]

A car is traveling at 51.0mi/h on a horizontal highway

If the coefficient of static friction between the road and tires on a rainy day is 0.097, what is the minimum distance in which the car will stop?

what is the stopping distance when the surface is dry and the coffefficent of static friction = 0.605?

The kinetic energy of the car is 1/2 mv^2. To get the car to a stop, you must do work on the car. The force that does the work is the friction force.

friction = f = u*N, where u is the friction coefficient, and N is the normal force of the object, i.e. the car. Here, N=mg, so f=umG.

The work done by this force is just w=f*d=umgd

Just set 0.5mv^2=umgd. m drops out, so

d=(v^2)/2ug

Then just plug-n-chug.

 

[thegymbum]

d=(v^2)/2ug

Then just plug-n-chug.

Well said 😉

Also don't forget units.. 51mi/h... 1mi = 1609m, 3600s = 1hour.

Have fun 😛