# Test tomorrow trouble with conversion.

[SOLVED] Test tomorrow... trouble with conversion.

Hi, I'm in Grade 11 Physics, and I have a Unit 1 Test tomorrow on motion and forces.

I'm not the smartest, so I can't exactly tell if the book is wrong. I'm doing a question out of the book, and was wondering what answers you guys come up with.

The question is: Calculate the net force acting on a 20kg object if the acceleration is:
a) 9.8m/s$$^{2}$$
b) 0.28m/s$$^{2}$$
c) 5669km/h$$^{2}$$
d) (50km/h)/s

Currently, I'm stuck on c). The book says the answer is 8.8N, and I'm getting 3160N. The calculated a value I got was 158m/s$$^{2}$$.

If I'm wrong, can you tell me how to correct myself?

Thanks a lot for your help, guys.

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1000 m = 1 km
1 h = 3600 s

F = m a

You forgot to use the 1 h = 3600 s conversion factor twice, that is, to square it. You have to do that because acceleration needs seconds squared in the denominator.

a = (5669 km/h^2) (1000 m/1 km) (1 h/3600 s) (1 h/3600 s)

or

a = (5669 km/h^2) (1000 m/1 km) (1 h/3600 s)^2

Last edited:
Thanks for the responses, but I'm still stuck here... What I've been doing is getting 5669, and dividing it by 60, then again by 60, getting me 1.57472, then I moved 3 decimals to the right, resulting in 1574.7. If I multiply that by 20 (m*a), I get 31494N. I'm still stuck. If someone can show me step how to actually solve this question, that'd be great.

a = (5669 km/h^2) (1000 m/1 km) (1 h/3600 s) (1 h/3600 s) = 0.437422839 m/s^2
F = ma = (20 kg)(0.437422839 m/s^2) = 8.74845679 N = 8.8 N

Last edited:
Thanks a lot mikelepore.

I actually just had it figured out thanks to you, right before I read your message. What I did though, was use 5669000kg/s^2 and divide it by 3600^2. That gave me the 0.43 number, and I was able to figure it out.

Thanks a lot. Your method confuses me a little though, because I'm just used to sliding decimal places instead of dealing with numbers. I'm still a little confused as to why I had to square the 3600. Thanks again.

EDIT: I assume we square the number because the seconds is already squared above, so we just add it in?

Reason you have to use the conversion factor twice is because units must obey the same laws of algebra that numbers do. See "dimensional analysis" in the index of your textbook.

If you did:
Code:
km       m          h
----    ----       ----
h^2     km          s
then the units will fail to cancel out to give exactly m/s^2 and nothing else.

But if you do:

Code:
km       m          h        h
----    ----       ----    ----
h^2     km          s        s
then the units cancel and all you are left with is m/s^2.

Ok, thanks a lot!

I'll tell you where most people make this particular type of mistake: in conversions related to area and volume. For example, suppose a certain liquid has a density of 1 g/cm^3. Let's convert that density to the SI unit of kg/m^3. To make the units cancel out properly, I'm going to have to use the conversion factor (100 cm / 1m) three times!

Code:
            g        1 kg       100 cm     100 cm    100 cm
1 ------    -------    --------   --------  --------
cm^3     1000 g       1 m        1 m       1 m

kg
= 1000 ----
m^3