Distance to Tow Auto to Reach Speed

Question:

A 2000 kg automobile is being towed up a 10 degree grade by a tow line making a 20 degree angle witn the road. The tension in the tow rope is 4000 N. How much distance must be covered in changing the speed from 2 m/s to 5 m/s? Neglect friction and the mass of the wheels. That means the rotational energy of the rolling parts need not be dealt with.

Answer:

Here the work-energy theorem may be applied. That says the change in kinetic energy of the auto is equal to the work done in accelerating it. The kinetic energy is 1/2*m*v² so its initial value was 0.5*2000*2² Joules = 4000 Joules. The final kinetic energy is 0.5*2000*5² = 25,000 Joules. The difference is 21,000 Joules.

The work done by the force applied through the tow rope goes into two things. Part of it accelerates the car and part of it raises the potential energy of the car by changing its elevation. The change in elevation as a function of distance, d, pulled is h=d*sin(10). The potential energy of the car then as a function of distance pulled is m*g*h or 2000 kg*9.8 m/s² *d m*sin(10) = 3410*d Joules. The total work, Wt done by the force attached to the rope is the tension in the rope times the distance moved times the cosine of the angle between the force vector and the displacement vector. Wt=4000 N*d*cos(20) = 3760*d Joules. The work which goes into accelerating the car is the difference between the total work and the lifting work, 3760*d-3410*d = 350*d Joules. When 350*d Joules = 21000 Joules, the car will have reached the required speed. So d = 21000/350 = 60 m.

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