How Do You Calculate the Sliding Distance with Friction and Initial Forces?

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The discussion focuses on calculating the sliding distance with friction and initial forces. The user successfully calculated the sliding distance after a running start but struggled with the second part involving a push on a cousin. They correctly identified the need to consider both the work done by the push force and the work done against friction. After some confusion, they realized the importance of calculating total work done during the push to determine the cousin's kinetic energy and subsequent sliding distance. The conversation highlights the application of physics principles in real-life scenarios.
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I got the first part done but I am 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 I am right there I just can't put the equation together because I am 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.
 
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Find her kinetic energy after the push. Hint: What's the total work done on her?
 
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??
 
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.
 
OHHH ok I got it finally ...appreciate all your help
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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