Question

A 3.7 kg block is being pulled up a rough incline, where θ= 21° and μk= 0.17, with an acceleration of 0.3 m/s/s. What is the magnitude of the Tension force pulling the block up the incline?

Answer 1

Step-by-step explanation:

Fn= normal force ( this determines the Ff friction)

Fn = mg cosΘ = 3.7 * 9.81 cos (21 ) = 33.88 N

Ff = 33.88 N * .17 = 5.76 N

Fdp = force downplane = mg sinΦ = 3.7 * 9.81 sin 21 = 13.01 N

Total forces acting down the plane = 13.01 + 5.76 = 18.77 N

T = The force acting up the plane to move and accelerate the block

Then :

F= ma

(T - 18.77 N) = 3.7 kg ( .3 m/s^2 )

T = 19.88 N