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m1 maths need help

a block of mass 6kg moves on a rough horizontal surface, coeff of friction 0.25, under the action of a horizontal force. It accelerates from rest to a speed of 4m s-1 in a distance 12m, continues for a time at this sped and then decelerates to rest in a distance of 2m. Find the magnitude and direction of the horizontal force requires during each stage of the journey.

Help please!

I've worked out stage 2, but stuck on stage 1 and stage 3

Step by step would be great thanx
The fact that the object is accelerating means there is a resultant force which is greater than the force of friction in this case.

u=0 v=4 s=12 a=?

With that information how would work out the acceleration?

There are only two forces that act in the horizontal; friction and your force. We know that the frictional force= uR so:

-> F - uR = ma

To solve this equation we need to work out what R is. Since there is no vertical acceleration, the downwards force(mg) is equal to the Reaction force.

With that knowledge, you have all you need to work out F.

For stage 3, the method is essentially the same except you will have a negative acceleration meaning that the Frictional force is greater than the horizontal force.
Reply 2
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svirdi
OP
I get the answer as 18.3 using your method for the first stage but book says 18.7????
Original post by svirdi
I get the answer as 18.3 using your method for the first stage but book says 18.7????



Must be rounding error. Try not to work out the exact value until the end. For example if you have a value of 2/3, keep it as 2/3 for further calculations rather than 0.67.

v^2= u^2 + 2as

4^2 = 2as
a = 16/24 = 2/3

F - 0.25R = 6(2/3) ----------------- R=6g therefore

F= 4 + (0.25 x 6 x 9.8)
F= 18.7
(edited 9 years ago)
Reply 4
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svirdi
OP
Thanx appreciate your help :smile:

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