Mechanics - Vertical Motion Questions

A question below - i've written how far i've got but i'm shooting in the dark a bit 'cos i'm not sure about it... any help appreciated and will be repped :biggrin:

A load of weight 7kN is being raised from rest with constant acceleration by a cable. After the load has been raised 20m, the cable suddenly becomes slack. The load continues upwards for a distance of 4m before coming to instantaneous rest.

(a) State the mass of the load

F=ma
m = f/a
m = 7000/acceleration

can i do this because it says constant acceleration?

(b) Draw a velocity time graph for the motion of the load. Use this to find the acceleration of the load before the cable becomes slack.

so the graph would go up steeply, then be less steep then stop? How do i find the acceleration from this?

(c) Assuming no air resistance find the tension in the cable before it comes slack

Don't know about this - is it to do with f=ma again? or t = mg?

Reply 1

Gaz031

Use energy considerations.
Energy is conserved.
KE = 0.5mv^2.
PE = mgh
EPE = (lamda)x^2/2l.

Reply 2

James.R

Gaz031

Use energy considerations.
Energy is conserved.
KE = 0.5mv^2.
PE = mgh
EPE = (lamda)x^2/2l.

Is that for M1? I don't think we've done that yet....only up to chapter 3 of M1.....sorry I should have said in the question

no, its M2.

for part a, Mass = 7000/9.8 = 714 kg (3 sf)

Reply 5

pkchips

for part b, the graph is triangle shaped. i dont know about using the graph, but heres one way I think of doing it:

It takes 4m for particle to come to rest,

v^2 = u^2 + 2as
Therefore, v^2 = 78.4

It took 20m to reach this velocity,

v^2 = u^2 + 2as
78.4 = 2 * a * 20
a = 1.96 ms-2