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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: jaydee on May 19, 2012, 11:11:31 pm

Title: Motion question
Post by: jaydee on May 19, 2012, 11:11:31 pm: A mass of 10kg is attached to an elastic cord with a stiffness of 100N/m. The mass is being released from the top of a cliff. The mass would come to rest momentarily when it is 20 m below the point of release. What is the speed of the mass when it is 18m below the initial point?
Title: Re: Motion question
Post by: Phy124 on May 20, 2012, 02:54:13 am: Quote from: jaydee on May 19, 2012, 11:11:31 pm
A mass of 10kg is attached to an elastic cord with a stiffness of 100N/m. The mass is being released from the top of a cliff. The mass would come to rest momentarily when it is 20 m below the point of release. What is the speed of the mass when it is 18m below the initial point?
Ok the way I'm going to do this using energy methods which involves 2 steps.

1. Finding the extension in the cord
2. Using this extension to find the velocity

*Note I will take my datum 20m below the top of the cliff (Note: a datum is like a reference point, e.g. the height in mgh at the datum equals 0)

Ok firstly, to find the extension in the cord.

$E_{top}=mgh= (10)(9.8)(20)$

^As it is not moving or being stretched

$E_{bottom}=\frac{1}{2}kx^2=\frac{1}{2}(100)x^2$

^As it is not moving and it is at the datum

Due to energy conservation these energies will be the same;

$E_{top}=E_{bottom}$

$(10)(9.8)(20)=\frac{1}{2}(100)x^2$

$x=\sqrt{\frac{(10)(9.8)(20)}{50}}\approx 6.26m$

If the extension at this point is 6.26 m, then the extension 2m above this point (where we want to find the velocity) will be 4.26 m

The energy at this point will consist of gravitational potential, kinetic and elastic potential energy.

$E_{2m \ above \ datum}=mgh+\frac{1}{2}mv^2+\frac{1}{2}kx^2=(10)(9.8)(2)+\frac{1}{2}(10)v^2+\frac{1}{2}(100)(4.26)^2$

Due to energy conservation $E_{2m \ above \ datum}=E_{top}$

$(10)(9.8)(2)+\frac{1}{2}(10)v^2+\frac{1}{2}(100)(4.26)^2=(10)(9.8)(20)$

$v=\sqrt{\frac{(10)(9.8)(20)-(10)(9.8)(2)-\frac{1}{2}(100)(4.26)^2}{(\frac{1}{2})(100)}}$

$v \approx 13.1 \ m/s$

Also, sorry it's a bit clustered I'm just using google calculator so didn't want to make too many mistakes so it was easier that way

edit: stupid slashes were the wrong way
Title: Re: Motion question
Post by: jaydee on May 20, 2012, 05:57:11 pm: the answer says 13.15m/s so i dunno :/
Title: Re: Motion question
Post by: Phy124 on May 20, 2012, 06:58:26 pm: See editing above.
Title: Re: Motion question
Post by: jaydee on May 20, 2012, 07:02:54 pm: thanks! i get it :)