Question

Help me please!!
At a playground, a 19.9-kg child plays on a slide that drops through a height of 2.31 m. The child starts at rest at the top of the slide. On the way down, the slide does a nonconservative work of -352 J on the child. What is the child's speed at the bottom of the slide?

Answer 1

so we will set it up like this:
PE_initial - 352J = KE_final
make sense?

Answer 2

Kind of. But where does the final speed come into play?

Answer 3

KE = (1/2) m v^2, it comes in the KE_final, do you see the 'V' in the equation, it stands for velocity, aka, speed

Answer 4

@jfisch1400 you still around?

Answer 5

yeah. what about PE_initial?

Answer 6

PE = mgh

Answer 7

so this:
PE_initial - 352J = KE_final
now looks like this:
mgh - 352 = (1/2)mv^2

you know mass, m, and
g, gravity
then just solve for v.

i can help with the algebra if you need

Answer 8

so, \[19.9*9.8*2.31-352=19.9v^2\] and i can do the algebra

Answer 9

i got 2.225

Answer 10

but that's not right.

Answer 11

don't forget the (1/2) on the right hand side of the equation, like this:
19.9∗9.8∗2.31−352=(1/2) 19.9 v^2

Answer 12

thank you very much. I have two others that I'm stuck on. could you help me with them?

Answer 13

yep, just post them up and I will help you!

Answer 14

I've been stuck on this one since last night.
An 8.57-kg block slides with an initial speed of 1.70 m/s up a ramp inclined at an angle of 26.6° with the horizontal. The coefficient of kinetic friction between the block and the ramp is 0.64. Use energy conservation to find the distance the block slides before coming to rest.

Answer 15

so this one we will have to use sin(26.6) , but setting up the equation it will look like this:
event 1 is the instant the block is sliding up at 1.7m/s
event 2 is the moment the block comes to a stop

PE + KE_1 -friction = KE + PE

(this one is ugly!)

Answer 16

PE_1 + KE_1 -friction = KE_2 + PE_2
mgh + (1/2)mv^2 - friction = (1/2)mv^2 + mgh

Answer 17

and friction = μmgcosΘ right?

Answer 18

but how do you get h_1 and h_2 for mgh?

Answer 19

so we will say that
PE_1 has h=0 becuase we are using this spot as our starting location.
PE_1 = 0
KE_1 = (1/2)mv^2, we know it's mass, and we know it's veloocity so
KE_1 = (1/2)8.57 * 1.7^2

Answer 20

"and friction = μmgcosΘ right?"
good, this is right, almost, inorder to use friction with KE and PE, we have to turn it into work, like this: Work = μmgcosΘ * distance

Answer 21

"but how do you get h_1 and h_2 for mgh?"
we set h_1 = 0, to use as our base elevation.
h_2, is where we will need to use sin(26.6)

Answer 22

this is tough stuff, what class is this? college?

Answer 23

AP physics, yes. and okay

Answer 24

see how event 1 and event 2 represent the corners of the triangle?

Answer 25

sin(26.6) = h_2/distance

Answer 26

this part is crucial to solving this problem, is it making sense?

Answer 27

now, this is where are are to now-
KE_1 - work = PE_2

where
KE_1 = (1/2)8.57 * 1.7^2
Work = μmgcosΘ * distance
PE_2 = mgh_2

Answer 28

brb, keep working on it, lemme know if you get stuck, but i think you are really close to getting this one!

Answer 29

PE_1=0
KE_1=1/2(8.57)1.7^2=12.384
friction=.64(8.57)(9.8)d=53.751d
KE_2=0
PE_2=(9.8(8.57))sin26.6=37.606
12.384-53.751d=37.606
-53.751d=25.222
d=.469
did I do evertything right? or did I miss anything?

Answer 30

let me see..
on friction don't for get the cos()
friction = .64*8.57*9.81*cos(26.6) * d
and on PE_2, the d term
PE_2 = 8.57*9.81*sin(26.6) * d

Answer 31

so friction = 53.751cos(26.6)d = 48.062d and the equation is
12.384-48.062d = 37.606d
12.384=85.668d
i got d = .145?

Answer 32

I believe in you! haha

Answer 33

yep, nicely done!

Answer 34

d = 0.145 meters for me too

Answer 35

you're the best. but I have one more if you don't mind?

Answer 36

haha, one more!, throw it up, i'm good for a few more minutes before my gf calls

Answer 37

An 81.0-kg in-line skater does +3550 J of nonconservative work by pushing against the ground with his skates. In addition, friction does -660 J of nonconservative work on the skater. The skater's initial and final speeds are 2.50 m/s and 1.60 m/s, respectively.
(a) Has the skater gone uphill, downhill, or remained at the same level? Explain.
(b) Calculate the change in height of the skater. (m)

Answer 38

alright, so same thing as before, set up our equation:
KE_1 + PE_1 + 3550j - 660j = KE_2 + PE_2

Answer 39

in general we have:
m = 81
g = 9.81

event 1 we have:
v_1 = 2.5

event 2 we have:
v_2 = 1.6

we need
h_1
h_2 (if they exisit)

Answer 40

ya?

Answer 41

so what i think we should do is assume either
h_1 = 0 or
h_2

and when we solve for the term that's not 0, we can see if we assumed right. we can figure we assumed right if the sign on the h we are solving for comes out correct. if we picked the wrong h, then we can know it's wrong becuase when solving for the other h, the sign will be wrong - like lets say h should be -h, but when we solve it we get +h

Answer 42

so if I get a positive h then it is uphill?

Answer 43

yaya, IF you assume h_1 = 0, and if h_2 becomes positive, it means it's up hill

Answer 44

sorry with the prior response, it was wordy!!

Answer 45

@jfisch1400 I gotta take off soon, is there any more i can help?

Answer 46

i got that h = .238 so he goes uphill? and this is the last problem i have. thank you for all the help!

Answer 47

I got 3.82 with this equation
.5*81*2.5^2 + 3550 -660 = .5*81*1.6^2 + 81 * h * 9.8

Answer 48

assuming h_1 = 0, and solve for h_2

Answer 49

oh I see what I did. Thank you so much!

Answer 50

you're welcome.
honestly, you are one of a few kids that actually learn, nice work!

Answer 51

thanks. I'm always open to learn new things in the pursuit of knowledge and happiness. do you teach physics?

Answer 52

I don't teach, but have been told that I should, huh.