
#1
Feb2409, 08:28 AM

P: 5

I got the first part done but im getting confused with B because I know the work done by the push  force by my friction should equal the force done by my cousins force by friction..I know Im right there I just cant put the equation together cuz Im getting stuck any help Please
Suppose that the coefficient of friction between your feet and the floor, while wearing socks, is 0.250. Knowing this, you decide to get a running start and then slide across the floor. a) If your speed is 3.00 m/s when you start to slide, what distance d will you slide before stopping? F*d = 1/2mv^2 .250m(9.8 )*d = 1/2m(3)^2 d = 1/2(3)^2 / .25(9.8 ) d = 1.8367 m b) Now, suppose that your young cousin sees you sliding and takes off her shoes so that she can slide as well (assume her socks have the same coefficient of friction as yours). Instead of getting a running start, she asks you to give her a push. So, you push her with a force of 125 N over a distance of 1.00 m. If her mass is 20.0 kg, what distance d does she slide (i.e., how far does she move after the push ends)? Remember that the friction force is acting anytime that she is moving. 



#2
Feb2409, 08:45 AM

Mentor
P: 40,906

Find her kinetic energy after the push. Hint: What's the total work done on her?




#3
Feb2409, 07:23 PM

P: 5

ok let me see so her kenetic energy afte rthe push will take the variables as t
m=20kg v=9.80 giving a energy of 960.4N TOTAL WORK DONE = is the work of 125N times the coefficient plus the 1.00m of distance now what?? 



#4
Feb2409, 07:47 PM

Mentor
P: 40,906

Need help to finish problem
To find the total work done during the push, consider the work done by the 125 N force and the (negative) work done by friction over the 1 m distance of the push.




#5
Feb2409, 09:13 PM

P: 5

OHHH ok I got it finally ...appreciate all your help



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