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physics acceleration resultant forces question
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M.H4kim
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#1
a 1500kg car accelerates at 1.5m/s^2 along a horizontal road.If the frictional forces acting against the car's motion total 1000N, what driving force is exerted on the road by the car's wheels?
pls help
I used f=ma to give total force as 2250N and subtracted 1000(frictional force) from this to give me 1250N as a driving force.
However this answer is wrong and I don't know what I am doing wrong.
pls help
I used f=ma to give total force as 2250N and subtracted 1000(frictional force) from this to give me 1250N as a driving force.
However this answer is wrong and I don't know what I am doing wrong.
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Muttley79
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#2
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#2
(Original post by M.H4kim)
a 1500kg car accelerates at 1.5m/s^2 along a horizontal road.If the frictional forces acting against the car's motion total 1000N, what driving force is exerted on the road by the car's wheels?
pls help
I used f=ma to give total force as 2250N and subtracted 1000(frictional force) from this to give me 1250N as a driving force.
However this answer is wrong and I don't know what I am doing wrong.
a 1500kg car accelerates at 1.5m/s^2 along a horizontal road.If the frictional forces acting against the car's motion total 1000N, what driving force is exerted on the road by the car's wheels?
pls help
I used f=ma to give total force as 2250N and subtracted 1000(frictional force) from this to give me 1250N as a driving force.
However this answer is wrong and I don't know what I am doing wrong.
Write out the equation of motion properly
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M.H4kim
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#3
(Original post by Muttley79)
Why have you subtracted 1000?
Write out the equation of motion properly
Why have you subtracted 1000?
Write out the equation of motion properly
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Muttley79
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#4
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#4
(Original post by M.H4kim)
F=ma?
F=ma?
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Muttley79
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#5
(Original post by rose.clm)
you
you
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M.H4kim
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#6
(Original post by rose.clm)
you would add on the 1000 frictional force, as the 2250 is the EXTRA force that is required to accelerate.
you would add on the 1000 frictional force, as the 2250 is the EXTRA force that is required to accelerate.
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M.H4kim
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#7
(Original post by M.H4kim)
I still don't get why u add 2250 and 1000
I still don't get why u add 2250 and 1000
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Muttley79
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#8
(Original post by M.H4kim)
because it is asking for just the driving force. Not the sum of the frictional forces and the driving force
because it is asking for just the driving force. Not the sum of the frictional forces and the driving force
Last edited by Muttley79; 3 months ago
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rose.clm
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#9
(Original post by Muttley79)
EDIT your post - you are breaking forum rules by giving an answer and you aren't helping the OP understand
EDIT your post - you are breaking forum rules by giving an answer and you aren't helping the OP understand
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Muttley79
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#10
(Original post by M.H4kim)
I still don't get why u add 2250 and 1000
I still don't get why u add 2250 and 1000
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