Calculate Distance of 5Kg Object on Frictional Surface: Matt's Help

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To calculate the distance traveled by a 5kg object on a frictional surface under a 15N force, the acceleration is determined to be 3 m/s². The final velocity is given as 12 m/s. Using the kinematic equation V² = 2ax, the distance can be calculated by rearranging it to x = V² / (2a). The initial calculations led to confusion regarding the time and distance, but the correct approach involves applying the appropriate kinematic formulas. The discussion emphasizes the importance of using the right equations to avoid errors in calculations.
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the problem is i have an object which has a mass of 5Kg it is on a frictional Surface a force of 15n acts on it which accelerates it to 12ms-1 i don't even know where to start to work out the distance i was thinking could you rearrange A= V-U divided by t (sorry it won't let me do the proper equation)


Into T= (V+U) x A
to work to work out time so i can put it into the Distance, Speed Time Triangle.
but
the answer 144 doesn't seem right am i doing it right or just going totality wrong ?

Thanks for your help in advance Matt
 
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Physics258744 said:
the problem is i have an object which has a mass of 5Kg it is on a frictional Surface a force of 15n acts on it which accelerates it to 12ms-1 i don't even know where to start to work out the distance i was thinking could you rearrange A= V-U divided by t (sorry it won't let me do the proper equation)


Into T= (V+U) x A
to work to work out time so i can put it into the Distance, Speed Time Triangle.
but
the answer 144 doesn't seem right am i doing it right or just going totality wrong ?

Thanks for your help in advance Matt

Welcome to PF.

Here are some equations that may address your interest in kinematics.

https://www.physicsforums.com/showpost.php?p=905663&postcount=2

From your force equation you know acceleration is 3m/s
Your final velocity is 12?

V2 = 2*a*x
 
Thank you :)
 
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