Question #99674
A railway car rolls down a small hill and comes to a level piece of track moving at 10 m/s. The coefficient of friction is 0.002 a) How much friction acts? b) at what rate does the car slow down? c) How far along the level track will it move before stopping? D) How much time does it take to stop
1
Expert's answer
2019-12-02T09:54:59-0500

a)


Ff=μmgF_f=\mu mg

b)


a=μg=(0.002)(9.8)=0.02ms2a=\mu g=(0.002)(9.8)=0.02\frac{m}{s^2}

c)


0.5mv2=mal0.5mv^2= mal

l=v22a=1022(0.02)=2500 ml=\frac{v^2}{2a}=\frac{10^2}{2(0.02)}=2500\ m

d)


t=va=100.02=500 st=\frac{v}{a}=\frac{10}{0.02}=500\ s


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