# our light truck weighs 3,500 kg. You drive up a gentle slope that is 0.5 km long.?

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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

• 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

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