You are Here: Home >< Physics

# Could someone help me with this Physics question please?

Announcements Posted on
Why bother with a post grad? Are they even worth it? Have your say! 26-10-2016

1. I think the value of 'h' is 0.1m from using sin(78)*2. I got 1.9 and I took this away from 2m.
I don't know how to do the second part, any help would be appreciated.
2. use mgh = 0.5mv^2 and rearrange for v
3. (Original post by DenizS)

I think the value of 'h' is 0.1m from using sin(78)*2. I got 1.9 and I took this away from 2m.
I don't know how to do the second part, any help would be appreciated.
You are on the right track, but need to be more accurate and round to three decimal places.

Once you have the height the pendulum falls through, plug that into mgh to get the initial potential energy.

Them rearrange the kinetic energy formula with v as the subject and using the value of mgh you calculated.
4. (Original post by uberteknik)
You are on the right track, but need to be more accurate and round to three decimal places.

Once you have the height the pendulum falls through, plug that into mgh to get the initial potential energy.

Them rearrange the kinetic energy formula with v as the subject and using the value of mgh you calculated.
I have rearranged the formula so it is v= the square root of 2mgh/m
I then substitute in the square root, 2(0.2*9.81*0.1)/0.2. This gives me 1.4.

Is 1.4 my initial velocity?
5. (Original post by DenizS)
I have rearranged the formula so it is v= the square root of 2mgh/m
I then substitute in the square root, 2(0.2*9.81*0.1)/0.2. This gives me 1.4.

Is 1.4 my initial velocity?

rearrange this to get v on it's own:

NB because m is on both sides of the equation it cancels out to leave:

now rearrange:

Your value of h using trigonometry is not accurate enough. Round this value to 2 decimal places (the question gave g = 9.81ms-2 so you need to be consistent with this.
6. (Original post by uberteknik)

rearrange this to get v on it's own:

NB because m is on both sides of the equation it cancels out to leave:

now rearrange:

Your value of h using trigonometry is not accurate enough. Round this value to 2 decimal places (the question gave g = 9.81ms-2 so you need to be consistent with this.
Okay, I substituted 0.04 as the change in height into the equation to get 0.7848. Is 0.78 my initial velocity?
7. (Original post by DenizS)
Okay, I substituted 0.04 as the change in height into the equation to get 0.7848. Is 0.78 my initial velocity?
You want to check your calculation again.
8. (Original post by DenizS)
Okay, I substituted 0.04 as the change in height into the equation to get 0.7848. Is 0.78 my initial velocity?

9. (Original post by uberteknik)

0.93
10. (Original post by DenizS)
0.93
Yes.

Maximum v is when all of the potential energy is converted to kinetic energy at the bottom of the swing.

v = 0.93 ms-1 (2 d.p.)
11. (Original post by uberteknik)
Yes.

Maximum v is when all of the potential energy is converted to kinetic energy at the bottom of the swing.

v = 0.93 ms-1 (2 d.p.)
Thank you

## Register

Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank
2. this can't be left blank
3. this can't be left blank

6 characters or longer with both numbers and letters is safer

4. this can't be left empty
1. Oops, you need to agree to our Ts&Cs to register

Updated: September 10, 2016
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Today on TSR

### University open days

Is it worth going? Find out here

### How to cope with a heavy workload

Poll
Study resources

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.