Will someone explain this physics tension problem to me?

Question

First, there's a diagram of a mass on an inclined plane. Attached to the mass is a rope that runs through the pulley at the top. The plane is essentially a right triangle.

The plane has is inclined at a 30-degree angle. The coefficient of kinetic friction between the mass and the plane is 0.1. The mass is 100 kg. If the mass is moving up the inclined plane at a velocity of 2 m/s, what tension should be applied to the rope in order to make the mass reverse directions in exactly 1 s?

a. 213 N
b. 387 N
c. 613 N
d. 787 N

I don't quite understand the question so please explain your reasoning. Thank you.

Answer 1

The equation for motion of the mass is
v(t)=vo+a*t
at the instant the mass changes direction, v(1)=0
0=2+a*1
a=-2

so f/m=a
f=-2*m

f is the net force on the block
to determine f, consider a FBD of the mass with the positive direction upslope

the forces parallel to the plane are
gravity
=-m*g*SIN(30)

and tension T

and friction, which is
=-N*u

N =cos(30)*m*g

sum the forces together and set equal to -2*m

T-m*g*SIN(30)
-COS(30)*m*g*u=-2*m

T=m*g*(sin(30)+cos(30)*u)
-2*m

using g=10 I get
387N

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