Mid Term Question
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The driver of a 2.0 × 103 kg red car traveling on the highway at 45m/s slams on his brakes to avoid striking a second yellow car in front of him, which had come to rest because of blocking ahead as shown in above Fig. After the brakes are applied, a constant friction force of 7.5 × 103 N acts on the car. Ignore air resistance.
Consider an automobile moving at v mph that skids d feet after its brakes lock. Calculate how far it would skid if it was moving at 2v and the brakes were locked.
Mass of Car = m = 2.0 x 103 kg
Velocity of Car = Vi = 45 m/s
Final Velocity = Vf = 0
Force of Friction = Fr = 7.5 x 103 kg
Least Distance = s = d = ?
As We know that
Ffr = mk
´ N = mk
.mg
Friction Force = Normal Force in Opposite Direction
Fn = -Ffr
mg = - mk
.mg
Acceleration = a = -mk
.g
We know that
g = 9.8 m/s2 and m
= 7.5 x 103 kg
Acceleration = a = 7.5 x 103 x 9.8
a = -73.5 x 103
According to Newton’s 3rd Law,
V2 = u2 + 2as ------------------------(1)
Put all values in equation (1)
0 = (45)2 + 2 (-73.5 x 103)
S = 0.0137m Ans
Physics, or the study of the elements, energy, and interactions between us helps us to understand the laws and laws that govern the universe. Not every student will grow up and learn physics at a deeper level, but everyone will use the basic concepts of physics to move around in daily life.
Consider an automobile moving at v mph that skids d feet after its brakes lock. Calculate how far it would skid if it was moving at 2v and the brakes were locked.
Vi = Vmph
First we convert it into mph and Feet
Vi = 0.44704 m/s
S = d feet
(0)2 = (0.44704)2 + 2a(0.30480)
0 = 0.1998 + 2 (0.30480)a
a = -0.3278
If Velocity is 2V, then
- = 2as
0 – (2)2 = 2 (-0.3278)s
S = 6.101m Ans
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