Asked by Angela
A 1500kg car skids to a halt on a wet road where μk= 0.53. How fast was the car traveling if it leaves 62-m-long skid marks?
Answers
Answered by
Phanor
First draw a free-body-diagram with Fn (normal force) pointing upward mg(Fg) pointing downward and the force of friction Ff to the left. Then set your axis for this problem I choice up and to the left as positive.
So you use Newtons 2nd law to find Fn
So the summation of Fy=ma
Since there is no motion in the y direction a=0
So it is Fg-ma so it's Fg=ma and it's (1500)(10) which gives you 15000N
Then you find Ff
Ff=u(mew)Fn
Ff=(.5)(15000)
Ff=7500N
Then you use F=ma because you are trying to find acceleration in the x direction which means you have to use the Ff because it is in the x direction so you get :
A=Ff/m
A=5 m/s^2
Then you need to find the initial velocity Vi so use
Vf^2=Vi^2 + 2ad
Vf=0 because it skids to rest so
0=Vi^2 + 2ad
-2ad=Vi^2
When you work all of it out you get -25.5 m/s it something around there
Hope this helped
So you use Newtons 2nd law to find Fn
So the summation of Fy=ma
Since there is no motion in the y direction a=0
So it is Fg-ma so it's Fg=ma and it's (1500)(10) which gives you 15000N
Then you find Ff
Ff=u(mew)Fn
Ff=(.5)(15000)
Ff=7500N
Then you use F=ma because you are trying to find acceleration in the x direction which means you have to use the Ff because it is in the x direction so you get :
A=Ff/m
A=5 m/s^2
Then you need to find the initial velocity Vi so use
Vf^2=Vi^2 + 2ad
Vf=0 because it skids to rest so
0=Vi^2 + 2ad
-2ad=Vi^2
When you work all of it out you get -25.5 m/s it something around there
Hope this helped
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