algebra in mechanics

[red and white kop!]

im not sure where to post this cos it includes mechanics formulae but im sure most of you have done mechanics as well since it is often included in maths programmes
so

a load of weight 7 kN is being raised from rest with constant acceleration by a cable. after the load has been raised 20 metres, the cable suddenly becomes slack. the load continues upwards for a distance of 4 metres before coming to instantaneous rest. Assuming no air resistance, find the tension in the cable before it became slack.

so obviously T = 7000 N + a(700), a being acceleration
but i can't seem to find this acceleration; ive tried using a velocity-time graph but there are too many unknowns for me; i am also a bit confused by what is meant by 'instantaneous' rest: does this mean constant velocity for 4 metres and then boom the graph line goes vertical?

 

[JeffWest]

You are on the right track. To find the acceleration we must split the problem in two. The first is when the load is being pulled by the cable with a constant acceleration and the second is when the cable is slack. When the cable is slack it is like there is nothing but gravity acting on it. It becomes a vertical projectile motion problem.

Your equation for tension is correct. Lets list the knowns

1. \(\displaystyle X_1=20 m\) change in position when accelerating
2. \(\displaystyle X_2=4\) m is the change in position when there is no tension
3. \(\displaystyle V_{o1}=0\) m/s there is no velocity to begin with
4. \(\displaystyle V_{f2}=0\) m/s there is no final velocity. (this it what it means by instantaneous rest. There is a split second that it stops moving, between it going up and when it will fall back down.)
5. \(\displaystyle V_{o2}=V_{f1}\) the final velocity of part one is equal to the initial velocity in part two.

There are things that we don't know...
1. t Time we do not know
2. a Acceleration due to the tension
3. \(\displaystyle V_{o2}=V_{f1}\) the velocity right when the tension is released.

When the tension is slack it will be good to use a constant acceleration equation.
The best one would be the one without time (because we do not know it).
The one that will work is \(\displaystyle V_f^2=V_o^2+2*a*X_2\)
we know that the final velocity is zero because it will be instantaneously at rest. we know the change in position \(\displaystyle X_2 = 4\), and the acceleration is just due to gravity because the tension is slacked.
Use this equation to find the Velocity right when the cable begins to slack. Use this as your final velocity of part one and use the same equation (\(\displaystyle V_f^2=V_o^2+2*a*X_2\)) to find the acceleration in part one.