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