Mechanical Energy Loss & Distance of Runner Slide

AI Thread Summary
The discussion revolves around calculating mechanical energy loss and the distance a runner slides into second base. The runner, weighing 70.0 kg and initially moving at 4.0 m/s, experiences a frictional force due to a coefficient of friction of 0.70. The mechanical energy lost to friction is determined to be -560 J. The normal force is calculated using the weight of the runner, leading to the equation for work done by friction. Ultimately, the distance slid is found to be approximately 1.16 meters, confirming the calculations are correct.
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Homework Statement


A 70.0 kg base runnner begins his slide into second base while moving at a speed of 4.0m/s. The coefficient of friction between his clothes and Earth is 0.70. He slides so that his speed is zero just as he reaches the base.

A) How much mechanical energy is lost due to friction acting on the runner? ( i solved A already)

B) How far does he slide?


Homework Equations


W=fdcos(0)


The Attempt at a Solution


I tried to take my answer from part A which was -560 J and have that equal the work, but I am stuck
 
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You have to find the frictional force.
What is the equation for that?
 
ff= ufn u= .70
 
Yes.
Now W = ff*d*(cos180), because frictional force acts in the opposite direction of the motion.
 
i don't have fn or the distance
 
fn=mu*N

then from W=fn*d you can get d.
 
fn = normal force = mg. You have already found W. Find d.
 
what would N be?
 
sorry, i had the wrong notation, I meant ff=mu*N and W=ff*d

Here I'm using N as the normal force or fn.
 
  • #10
so ff x d x (cos180) = -560

480.69 x d x cos 180= -560

d= 1.16??
 
  • #11
looks good to me.
 
  • #12
Your answer is correct.
 
  • #13
thank you for your help!
 
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