# Tension of a cord

1. Homework Statement
A 4.0 kg block is released from rest at the top of a frictionless plane of length 8.0m that is inclined at an angle of 15 degrees to the horizontal. A cord is attached to the block and trails along behind (assume the cord is massless). When the block reaches a point 5.0m along the incline from the top, someone grasps the cord and exerts a constant tension parallel to the incline. The tension is such that the block just comes to rest when it reaches the bottom of the incline. What is this constant tension? Solve the problem twice, once using work and energy and again using Newton's laws and constant acceleration kinematics.

2. Homework Equations
(1) dfy = diy + viyt - 1/2gt^2
(2) dfx = dix + vixt +1/2at^2
(3) Fn - mgcos15 = 0
(4) mgsin15 - Ft = max

3. The Attempt at a Solution
I'm really confused with this one. I've only attempted to solve the problem with Newton's laws and acceleration kinematics so far.
I used the first equation above to solve for time t = .5139s. Then I used the second equation to solve for ax = 18.796 m/s^2.
Lastly, I used the fourth equation to solve for Ft = -65N.
Can anyone confirm if this is correct? If it isn't, where did I go wrong?

Also, I don't know how to solve using work and energy. I know that work of a nonconservative force = change in kinetic energy + change in potential energy, but I don't know where to go from there.
1. Homework Statement

2. Homework Equations

3. The Attempt at a Solution