Answer to Question #250775 in Classical Mechanics for Mr Amazing

Question

A tractive effort of 3kN drives a 5 ton truck from rest up an inclince of 1 in 40. If the tractive resistance is constant at 380N and the tractive effort is parralel to the plane of the incline, use the method of conservation of energy to calculate the speed of the trcuk after it has traveled 250m up the incline.

Clearly show the following values:

Ek=?

Ep=?

ETR=?

ETE=?

Energy equation=?

v=?

Answer 1

$m =5t = 5000kg$

$F_{te}= 3kN=3000N$

$F_{tr}=380 N$

$l=250m$

$\frac {l}{h}=\frac{1}{40}$

$g = 9.8\frac{m}{s^2}$

$\text{Solution:}$

$E_k =\frac{mv^2}{2}= \frac{5000*v^2}{2}=2500v^2J$

$h = l*\frac{1}{40}= 250*\frac{1}{40}=6.25m$

$E_p = mgh = 5000*9.8*6.25= 306250 J$

$E_{tr}= F_{tr}*l = 380*250= 95000J$

$E_{te}= F_{te}*l = 3000*250= 750000J$

$E_{te}-E_{tr} = E_p+E_k$

$750000- 95000= 306250+2500v^2$

$v \approx 11.81\frac{m}{s^2}$