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.
PLEASE ANSWER ACCORDINGLY this is the third time im reposting because answers are either wrong or irrelevant
Clearly show the following values:
Ek=?
Ep=?
ETR=?
ETE=?
Energy equation=?
v=?
Consider the diagram
Force "\\times" Displacement "=\\frac{1}{2}v^2+\\frac{1}{2}u^2"
The net force on the track
"F_{net}= 3000N-380N-5000*9.81* sin (1.432)=1394.215 N"
Acceleration
"a_{net}=\\frac{F_{net}}{m}=\\frac{1394.215}{5000}=0.279 m\/s^2"
Assuming the force starts at rest
"v^2=u^2+2as\\\\\nv^2=2as\\\\\nv=\\sqrt{2as}\\\\\nv=\\sqrt{2*0.279*250}\\\\\nv=11.811 m\/s"
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