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

NetherCraft 0

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

Also Check This  why do people keep saying asians are victims of racism when all the asians i know are well off and responsible ?

Leave a Reply

Your email address will not be published. Required fields are marked *