A 1000 kg car is traveling at 40 m s-1 when its brakes are suddenly applied. The wheels
lock and the car skids to rest. If the frictional force between the tyres and the road is
6000 N, how far does the car skid before coming to rest?
A 6.7 × 10-1 m
B 6.7 m
C 1.3 × 102 m
D 2.4 × 102 m
E 2.7 × 102 m
F 7.5 × 102 m
G 1.5 × 103 m
H 2.0 × 103 m
Frictional force
2008-01-17 3:42 pm
回答 (3)
2008-01-17 5:24 pm
✔ 最佳答案
F=6000N m=1000kg u=40ms^-1 v=0Since F=ma
the deceleration of the car is 6000/1000=6ms^-2
Applying v^2-u^2=2as
0^2-40^2=2(-6)s
-1600=-12s
s=133.33m
The nearest answer is C
2008-01-22 7:52 am
Now m=1000kg, v=40m s-1
KE(Kinetic energy) stored by the car
=1/2(m)(v^2)
=1/2(1000)(40^2)
=800000(J)
When a force of 1N moves 1m( in the direction of force), 1J of work is done.
The equation is:
Workdone(W)=Force in the dirction of displacement(F) x Displacement(s)
Now the stored energy of the car is 800000 J, the frictional force is 6000 N
So
800000=6000 x s
s=133.33333
So the nearest answer is C 1.3 x 102m
2008-01-21 23:56:48 補充：
呢個方法係用conservation of energy來做。將car的kinetic energy 轉做 work done of the car due to friction.
KE(Kinetic energy) stored by the car
=1/2(m)(v^2)
=1/2(1000)(40^2)
=800000(J)
When a force of 1N moves 1m( in the direction of force), 1J of work is done.
The equation is:
Workdone(W)=Force in the dirction of displacement(F) x Displacement(s)
Now the stored energy of the car is 800000 J, the frictional force is 6000 N
So
800000=6000 x s
s=133.33333
So the nearest answer is C 1.3 x 102m
2008-01-21 23:56:48 補充：
呢個方法係用conservation of energy來做。將car的kinetic energy 轉做 work done of the car due to friction.
參考: My phy knowledge
2008-01-18 1:35 am
答案是A,靠估
收錄日期: 2021-04-13 14:58:26
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