a 1200 kg car going 30 m/s applies its brakes and skids torest. if the friction

Question

a 1200 kg car going 30 m/s applies its brakes and skids torest. if the friction force between the sliding tires and thepavement is 6000N, how far does the car skid before coming to rest? I know the answer is 90m.
I have the one formula.... KE= 1/2 m (v^2)....... I don'tknow the other formula..... where i divide the 6000N
Can you help me with the other formula....... I know you divide butwhat should the formula(s) actually look like ?   thanksin advance .

KE = 1/2 m (v^2)
KE= 1/2 (1200) (30^2)
KE = 54000
now i divide 5400 by 6000 and come out to 90m

Explanation / Answer

As done by you , Kinetic energy = KE = 1/2 m (v^2) KE= 1/2 (1200) (30^2) KE = 54000 J When friction force is acting , it is actually doing workagainst the motion. Now, let the displacement is d Then work done = force*displacement = 6000*d As finally , the car stops , So final energy is zero . So initial kinetic energy =work done 54000=6000*d => d=90 m

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