Question

1. A 4500 kg ore car accelerates at 0.4 m/s² up a ramp which has a 9 % grade. If the coefficient of friction µ_k = 0.25, calculate the force applied on the ore car. (4 points)

please type your answer, Please do not write on the paper and post it

Answer 1

R = normal force on the body by the surface of incline

m = mass of the body

θ = angle of incline

μ = coefficient of kinetic friction

a = acceleration of the body

F = kinetic frictional force

perpendicular to incline surface , force equation can be given as

R = mg Cosθ                                  eq-1

kinetic frictional force is given as

F = μ * R

using eq-1

F = μ mg Cosθ                              eq-2

parallel to incline , force equation is given as

mg Sinθ - F = ma

using eq-2

mg Sinθ - μ mg Cosθ = ma

a = g (Sinθ - μ Cosθ )

Using g = 9.8

a = 9.8 (Sin35 - 0.32 Cos35 ) = 3.05 m/s2

Using Kinematic Equation

V^2 - U^2 = 2aS

U = 0

V = sqrt(2aS) = sqrt(2*3.05*45) = 16.57 m/s

Please Upvote if you like the answer!!