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posted by  RN on 5/20/2008 11:19:31 AM  |  status: Live  

acceleration

Course Textbook Chapter Problem
General Physics N/A N/A N/A
Question Details:
A force accelerates a 5 kg box from a speed of 2.5 m/s to a speed of 4.0 m/s in 5s.  The work done is by this force over this time is nearly equal to:
 
a) 25J
b) 47J
c) 62.5J
d) more information is needed
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Scholar
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posted by ProfF on 5/20/2008 11:37:18 AM  |  status: Live
Asker's Rating: Helpful   
RN's comment:
"thank you"
Response Details:
Let's start by determining the force applied, since we know that Work = F*d.
 
We can determine the force applied from Newton's 2nd Law: F=ma.  We know the mass, but we don't know the acceleration, so we can use kinematics to determine our acceleration:
vi=2.5 m/s
vf=4 m/s
Δt=5s
a=?
 
Use vf=vi+at ==> a=(vf-vi)/t = (4m/s - 2.5 m/s)/5s = 0.3 m/s^2
 
Now we can determine the force applied: F=ma = 5kg*0.3m/s^2 = 1.5 kg*m/s^2 = 1.5 N
 
Finally, we need the distance over which this force was applied, d.  Let's find this by going back to our kinematic equations:
vi=2.5 m/s
vf=4 m/s
a=0.3 m/s^2
Δt=5s
Δx=?
 
We can get this from vf2=vi2+2aΔx, which, solving for Δx, gives us 16.25m
 
Now, calculating Work = F*d, we have 1.5 N * 16.25m ˜ 24.4 N*m = 24.4 J
Scholar
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posted by zhsizh on 5/20/2008 11:44:58 AM  |  status: Live
Asker's Rating: Helpful   
RN's comment:
"thank you"
Response Details:
a) is correct!
 
a = (v1-v0)/t = (4.0m/s - 2.5m/s)/5s = 0.3m/s2
S = v0t + 1/2*a*t2 = 2.5m/s*5s + 1/2*0.3m/s2 *(5s)2 = 16.25m
F = ma = 5kg*0.3m/s2  = 1.5N
W = FS = 1.5N*16.25m = 24.375J
Apprentice
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posted by MSO on 5/20/2008 12:07:18 PM  |  status: Live
Asker's Rating: Helpful   
RN's comment:
"thank you"
Response Details:
intial velocity = 2.5 m/s
final velocity = 4 m/s
time = 5 s
 v= vi + a t    ------                              4 = 2.5 + 5 a
the acceleration = 0.3 m/s2
F=ma
F = 5 ( 0.3) = 1.5 N   
     Δx =vi t + 0.5 a t2
Δx = 2.5(5)+0.5(0.3)(5)2
Δx= 16.25 m
W = F Δx = (1.5)(16.25) = 24.375 ˜ 25  J
Oracle
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posted by zsm28 on 5/20/2008 12:17:57 PM  |  status: Live
Asker's Rating: Helpful   
RN's comment:
"thank you"
Response Details:
the work done = change of kinetic energy = final kinetic energy - initial kinetic energy
= mvf2/2 - mvi2/2 = m(vf2 - vi2)/2 = 5*(4.02 - 2.52)/2 = 25 J
Guru
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posted by Pradeep on 5/20/2008 2:16:08 PM  |  status: Live
Asker's Rating: Helpful   
RN's comment:
"thank you"
Response Details:
Given :

Mass of box, m = 5kg
initial velocity , u =2.5m/s
final velocity, v = 4m/s

to find : work done

By energy equation

Work done = change in kinetic energy = m(v2 - u2)/2 = 5(16 - 6.25)/2 = 24.375 J  ˜ 25J

so option A is correct
Guru
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posted by John Walker on 7/1/2008 2:18:02 PM  |  status: Live
Asker's Rating: Somewhat Helpful   
RN's comment:
"i rate higher for early responses thx"
Response Details:
a) 25J
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