# Kinetics of Particles work and energy; A moving car

by sharpe_116
Tags: energy, kinetics, moving, particles, work
 P: 1 1. The problem statement, all variables and given/known data The 2Mg (i assumed mega-gram, not sure if that is the correct term) car has a velocity of v= 100km/h when the driver sees an obstacle in front of the car. If it takes 0.75 s for him to react and lock the brakes, causing the car to skid, determine the distance the car travels before it stops. The coefficient of kinetic friction between the tires and the road is Uk= 0.25. 2. Relevant equations Work and energy for a system of particles: $$\Sigma$$T1 + $$\Sigma$$U = $$\Sigma$$T2 T's represent initial and final kinetic energy respectively, 1/2*m*v^2 U represents all work done by external and internal forces acting on the system. Work of a constant force along a straight line: U = Fcos$$\vartheta$$($$\Delta$$s) 3. The attempt at a solution So i tried to apply my basic equation to the question $$\Sigma$$T1 + $$\Sigma$$U = $$\Sigma$$T2 T2 equals zero (i assumed) because the car will come to rest at the end of the question T1 will equal 1/2 mv^2 which is equal to 1/2 *(2Mg * 100km/h^2) = 10000 since the only work acting in this question is the friction force Caused by the car braking U = Fcos$$\vartheta$$($$\Delta$$s) therefore U = Ff(force of friction)*$$\Delta$$s which is = -4.905$$\Delta$$s (negative because friction force acts in the negative direction) so to summarize i now have 10000 -4905$$\Delta$$s = 0. i solved for delta s and got 2038, but this is obviously incorrect, i dont know how to apply the time delay into this question. For reference the correct answer is s = 178m I just started this chapter and dont have my bearings yet so if you could please explain clearly how to proceed it would be appreciated, thank you Attached Thumbnails
 P: 23 Without doing any of my own work, here are a few pointers: Be sure to account the fact that the given velocity is in km/h and not m/s. Remember that the work done by a constant force can be written as F*x where x is the distance along which the force did the work. Now, to account for the time delay, simply find the distance the car would have traveled in that time and add it to the final distance to find the distance the driver travels before stopping.
 P: 460 use equations of motion to get this one

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