A 1200 kg car is moving in a straight line at a constant speed of 15 m/s when the brakes are applied. The coefficient of friction between the car and the road is 0.25. How far in m does the car travel after the brakes are applied?
Podobne:
- To Hire Travel Agency On Visa Application Will Give Better Chance Of Approval Than Doing It On Your Own? we applied US tourist visa last july with my two sons, but was denied, (the consular officer did not even bother to see our requirements, it was very unfair) i......
- To Hire Travel Agency On Visa Application Will Give Better Chance Of Approval Than Doing It On Your Own? we applied US tourist visa last july with my two sons, but was denied, (the consular officer did not even bother to see our requirements, it was very unfair) i......
- I`m Only Sixteen Years Old But I Want To Travel. What Are Inexpensive And Safe Ways For Me To Travel? I know I am a little young, but I want to travel more than anything and it can`t wait. Im working two part time jobs whenever I can to make......
because there is Friction involve, work is done
Work = ΔEnergy
Work = PEf – PEi
Intially, the car has Kinetic Energy, when it stops, its final energy is 0J, because work takes out of its energy
Work = 0 – KEi
Work = -KE
we know that Work = Force * distance. In this case, the force is the Firction force
Friction * d = – KE
Friction = μ(Fnormal)
the normal force in this case is equal to the weight
Friction = μ(weight)
weight = mg
Friction = μ(mg)
again, Friction * d = -KE
μ(mg) * d = -.5mv^2
the mass cancel out, leaving you with:
μg * d = -.5v^2
.25(-9.8) * d = -.5(15)^2
-2.45d = -112.5
d = 45.92m
The data is not sufficient to tell. The friction limits the deceleration to some value not exceeding 0.25 g, or 9.8 m/sec^2, but the brakes will probably not (and definitely should not) apply that much deceleration lest there be a skid. Since the actual deceleration will be less, a minimum distance can be determined, but the actual distance will be more. The mass of the car is irrelevant.
The friction force acting against the car’s movement is:
F=(1200)(9.8)(0.25)
the acceleration, opposed to movement, is:
a=F/m = (9.8)(0.25)
the distance is given by:
d = (vf^2-v0^2)/2a
where vf is the final velocity (0) and v0 is the initial velocity (15 m/s)
do the math.