How Do You Calculate Speed at the Bottom of an Inclined Plane?

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To calculate the speed at the bottom of an inclined plane, the kinetic energy equation KE = 1/2 mv^2 is used. By rearranging the equation to isolate v, one can take the square root after manipulating the terms. Specifically, if KE is 281 J and mass m is 5 kg, the calculation leads to v = √(2*281/5). This results in a final speed of approximately 12.57 m/s at the bottom of the incline. Understanding how to manipulate the equation and apply square roots is essential for solving such physics problems.
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If the mass slides down the inclined plane fomr the top, what will the speed be when it reaches the bottom

so my equation is KE=1/2mv^2

so 281 J = 1/2 (5) v^2

Im blanking out on how to cancel out the square aaah thx


EDIT : I GOT IT THanks
 
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You've done the physics, don't let the arithmetic stop you now!
To "cancel" an operation, use the opposite operation. What is the opposite of squaring?

Try using a square root!

If (5/2)v2= 281 then v2= 562/5. Take the square root of both sides.
 


No problem! To solve for v, you can rearrange the equation to isolate v. So, you can divide both sides by (1/2) and take the square root of both sides to cancel out the square. This will leave you with v = √(2KE/m). Plugging in the values, you would get v = √(2*281/5) = 12.57 m/s. Hope that helps!
 
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