Truck brakes can fail if they get too hot. In some mountainous areas, ramps of loose gravel are constructed to stop runaway trucks that have lost their brakes. The combination of a slight upward slope and a large coefficient of rolling friction as the truck tires sink into the gravel brings the truck safely to a halt. Suppose a gravel ramp slopes upward at 6.0∘ and the coefficient of rolling friction is 0.49.How long the ramp should be to stop a truck of 15000 kg having a speed of 35 m/s.
Taking work done by friction to be negative I get the following equations:
$K={1\over2}mv^2$
$W = Fd$
${1\over2}mv^2-F_f*d=0$
With the force diagram above I can now write $F_f = {\mu}mg_y = (.49)(15000)(9.81)cos(6^{\circ})$.
Then I just need to solve for $d$ in my work/energy equation. Doing so, I get
$${1\over2}mv^2-d({\mu}{mg}\cos(6^{\circ}))=0$$
$${1\over2}mv^2=d({\mu}{mg}\cos(6^{\circ}))$$
$$v^2=2d({\mu}{g}\cos(6^{\circ}))$$
$${{v^2}\over{{2\mu}g\cos(6^{\circ})}}=d$$
$${{35^2}\over{(2)(.49)(9.81)\cos(6^{\circ})}}=d$$
$$d=128m$$
This is not the right answer according to mastering physics.
Best Answer
HINT:
You must have to subtract the component of weight working on the same direction as the friction force. The equation shold be $$E_k-(F_f+W_x)d=0$$ Where,$W_x=mg\sin(6°)$