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Old Jan 26th 2009, 05:15 AM   #1
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Question on accelaration 2 (pls guide me,tq)

A 40-kg girl and a 8.4-kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope the girl exerts a 5.2 N force on the sled pulling it toward her.

a. What is the acceleration of the sled?
b. What is the acceleration of the girl?
c. How far from the girl’s initial position do they meet presuming the force remains constant and the no frictional forces act?

Here is my working,am i right?Pls guide me.Thank you.

a.the acceleration of the sled = a,
F = m1 * a
a = (F/m1)
F = 5.2 N
a = (5.2/40)
= 0.13 m/s2

b.the acceleration of the girl = a1,
F = m2 * a1
a1= (F/m2) = (5.2/8.4)
= 0.62 m/s2

c.The distance covered by the sled when the girl pulls it is
S = ut + (1/2)at2
S = 15 m,u = 0 m/s and a = 0.13 m/s2
15 = 0 * t + (1/2) * 0.13 * t2
t2= (15/0.065) = (15000/65)
t = 15.2s
The distance from the girl's initial position when they meet is
S1 = ut + (1/2)a1t2
u = 0 m/s
S1 = 0 * 15.2 + (1/2) * 0.62 * (15.2)2
S1 = 71.6 m

Last edited by jonbrutal; Jan 26th 2009 at 05:45 AM.
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Old Jan 26th 2009, 07:04 PM   #2
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For the acceleration in part a) and b), it is better to have one being positive and one being negative as they are in opposite directions.

For part c), you should note that both the girl and the sled are moving so it is not correct for you to find the time be at a distance of 15m. You should let the distance from the girl's position where they met be x m and on the other hand the distance from the sled's position where they met would be 15-x m, then you can use equation of motion and solve for x

Last edited by werehk; Jan 26th 2009 at 07:07 PM.
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Old Jan 26th 2009, 08:29 PM   #3
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Is this what you meant?

I tried what u said,should i find the time first?
This is what i did,

The distance of the girl from initial position
x = 1/2 (o.62)t*2=0.31t*2

the distance of the sled from the initial position
15 - x = 1/2(0.13)t*2=0.065t*2

SHould i equal both equations and find the value of x??Like this

x = 0.31t*2
t*2= x/0.31

15-x=0.065t*2
t*2= (15-x)/0.065

(15-x)/0.065 = x/0.31
0.065x=4.65-0.31x
x = 12.4 m


Please help to explain,thank you for your help & support.I will make the accelaration of the sled negative.

Last edited by jonbrutal; Jan 26th 2009 at 09:18 PM.
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