14000 obviously.!

1) F = Mg SIN(A)

2) R = Mg COS(A)

Note: If there was friction the equations would become

1) F = Mg SIN(A) UR (U = coefficient of Friction)

2) R = Mg COS(A)

You need to resolve forces along the slope, and this requires using the angle of the slope as well as the weight of the truckessentially you get 2 equations

M = Mass of Truck, g = Gravity, A= angle of slope, R = reaction of Ground to truck, F = Force of engine

1) F = Mg SIN(A)

2) R = Mg COS(A)

1) F = Mg SIN(A)

2) R = Mg COS(A)

14000 obviously.!

so F = 3500 x 10 x 0.4 = 14000N

therefore

The force exerted by engine = 14000N

Note: If there was friction the equations would become

1) F = Mg SIN(A) UR (U = coefficient of Friction)

2) R = Mg COS(A)

W = F * S

W = maS

work done is against gravity .

W = mgh

take g = 10

W =3500*10*200

=7000000

W = FS

F = 7000000/500 =14000N

Source(s): brain

12000N

so F = 3500 x 10 x 0.4 = 14000N

therefore

The force exerted by engine = 14000N

This post is last updated on hrtanswers.com at Date : 1st of September – 2022