Your light truck weighs 3,500 kg. You drive up a gentle slope that is 0.5 km long. At the top of the slope you are 200 m higher in elevation than when you started. Assuming no friction, what force has your engine exerted to drive up this slope?
A. 7000N
B. 12000N
C. 14000N
D. 20000N
2 Answers

the slope (a.k.a the triangle’s hypotenus = 0.5 km = 500m)
the height travelle (aka the triangle’s opposite = 200m)
it follow that, using thereom phythagoras, that sin Q = O/H
(I use Q as theta)
sin Q = 200/500
Q = 23.6 degrees
weight of the truck = W = mg = 3,500g
(g for gravitational force, some use the value 9.81 and some uses 10)
Therefore, the force exerted by engine
= W sin Q
= 3,500 x g x sin 23.6 degrees
= 14000N ( Iused g as 10 m/s^2 for this case)

14000 obviously.!
Source(s): expert