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*v^{2} so its
initial value was 0.5*2000*2^{2} Joules
= 4000 Joules. The final kinetic energy is 0.5*2000*5^{2} = 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/s^{2} *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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