Mechanics Newton's 2nd law question

[pianogirl]

Homework Statement

A load of 200kg is lifted through 50m by a vertical cable. The load accelerates uniformly from rest for 20 seconds, travels at constant speed for 10 seconds then decelerates to rest in a further 10 seconds. Find the maximum speed attained and the tension in the cable at each stage.

Homework Equations

F=ma and maybe some SUVAT ones?

The Attempt at a Solution

I started using F=ma so that for the first stage I got T=200a+1960 but soon after that I got a bit confused.

Help would be really appreciated because I'm stuck!
Thanks!

 

[tiny-tim]

Hi pianogirl!

Forget the tension … you need to find the acceleration before you can find the tension.

Write out the standard constant acceleration equations for the three stages of motion, plus the equation that says the three distances add to 50m.

What do you get? ​

 

[kerimek]

Hello! It looks like you're on the right track with using Newton's second law (F=ma) to solve this problem. Let's break down the problem into its three stages and use the appropriate equations to find the maximum speed and tension in the cable at each stage.

First Stage (Acceleration):
Since the load is accelerating uniformly from rest, we can use the SUVAT equation v=u+at, where v is the final velocity, u is the initial velocity (which is 0 in this case), a is the acceleration, and t is the time. We know that the load travels 50m in 20 seconds, so we can plug in these values to get:

v = 0 + a(20)
50 = 0.5a(20)^2
a = 0.5 m/s^2

Now, we can use F=ma to find the tension in the cable at this stage:

T = 200(0.5)
T = 100 N

Second Stage (Constant Speed):
Since the load is travelling at a constant speed, we know that the acceleration is 0. Therefore, we can use the equation v=u+at again, but this time, a=0. We also know that the load travels at this constant speed for 10 seconds, so we can plug in these values to get:

v = 0 + 0(10)
v = 0 m/s

This means that the maximum speed attained at this stage is 0 m/s, since the load is travelling at a constant speed and does not change its velocity.

Third Stage (Deceleration):
Finally, the load decelerates from this constant speed to rest in 10 seconds. This means that the acceleration is again non-zero, and we can use the SUVAT equation v=u+at to solve for the final velocity:

v = 0 + a(10)
0 = v + a(10)^2
a = -0.1 m/s^2

We can now use F=ma to find the tension in the cable at this stage:

T = 200(-0.1)
T = -20 N

Note that the tension is negative because it is acting in the opposite direction of the acceleration.

So, to summarize:
Maximum speed attained: 0 m/s
Tension in the cable during acceleration: 100 N
Tension in the cable during deceleration: -20