An ice-skater pushes heself away from a wall. She slides on the ice for a distance before stopping. A) has the ice-skater done work when she pushed herself? B) friction is a force. Does friction di work on the ice-skater? Give reasons for you answer
1. An ice-skater pushes heself away from a wall. She slides on the ice for a distance before stopping. A) has the ice-skater done work when she pushed herself? B) friction is a force. Does friction di work on the ice-skater? Give reasons for you answer
jawab
Down vote
I agree with CuriousOne that the example is more confusing than helpful, but this is the way I would explain it.
Suppose you take a spring, place it on the ground then compress it. If you now suddenly let go of the spring it will rebound and bounce upwards off the ground:
Spring
The spring clearly has work done on it because its kinetic energy increases and that increase must have come from somewhere. However the ground can't have done any work on the spring because the ground hasn't moved. It should be obvious that the potential energy in the compressed spring has been converted into kinetic energy of the uncompressed spring - in effect the spring has done work on itself. This is what your book means by:
transfers of energy from one type to another inside the object
i.e. potential energy of the compressed spring has been converted into kinetic energy of the uncompressed spring.
In the case of the skater the skater's arms correspond to the spring and the rail corresponds to the ground. The skater's arm isn't a spring, of course, because it's chemical energy not potential energy being converted to kinetic energy by the skater's muscles. Nevertheless the same principle applies.
Share Improve this answer Follow
answered
Oct 17 '14 at 9:29
John Rennie
319k●100100 gold badges●667667 silver badges●938938 bronze badges edited
Oct 17 '14 at 14:16
Up vote
4
Down vote
I think this is a good example, and should be studied and understood carefully. But I don't know where the book goes after making it. Whether or not it is poorly stated depends on the surrounding context.
It focuses attention on two things:
1.) the skater is a deformable body. The center of mass is not fixed in the body.
2.) work is defined as a force times the displacement of the point of application.
(And, arguably a third aspect: the skater carries a store of chemical energy. But this fact is aside the point being made.)
In this case, the point of application does not move. There is no external work from that force. Both of these points are important to understand and remember. Typically, up to this point in a course, problems are implicitly formulated in terms of a point-particle model of an object. No internal structure. When deformable bodies are considered, certain things that work for point-particles no longer work. Additional analysis is needed.
Note that there is internal work being done. Chemical reactions contract muscle fibers. The fibers apply forces to tendons. Now there is a force and a displacement of the application of the force.
Share Improve this answer Follow
answered
Oct 17 '14 at 12:43
garyp
20.8k●11 gold badge●3838 silver badges●7979 bronze badges
So, with your point 2), you mean to say that when a car crashes into a tree, the tree doesn't do any work on the car because the point of application of its force is fixed? –
gatsu
Dec 17 '15 at 6:41
Absolutely correct. –
garyp
Sep 25 '16 at 2:13
Add a comment
Up vote
3
Down vote
Accepted
An initially ice-skater pushes away from a railing and then slides over the ice. Her kinetic energy increases because of an external force F⃗ on her from the rail . However, that force does not transfer energy from the rail to her. Thus, the force does no work on her.
This is simply confused: This example is nothing but an elastic collision between the girl and the rail. The example given by John is appropriate but I do not agree with him that the arms behave differently from a spring. Of course the ultimate source of energy is chemical, but the girl as a system behaves exactly like a steel spring or a football hitting the rail.
enter image description here
If the skater (m = 30 Kg) had a spring and was travelling at v = 4 m/s she would have KE = 240 J and would bounce back at (roughly) the same speed. If she stretches her arms and pushes at the rail (like a spring) she can acquire the same speed of 4 m/s producing net work of 240 J. Because of human muscle inefficiency she'll have to burn 4-5 times that amount of calories $\approx 1080 J$
The body is compressed against the rail like an elastic ball, then it stretches back (in a ball the whole body, in a skater the arms, in a swimmer in a pool or in a high-jumper the legs) giving acceleration to the whole body. I
2. A skater moves with 15 m/s in a circle of radius 30 m . The ice exert a central force of 450 N. What is the mass of the skater. (GAFSA)
Jawaban:
BANYAK BANGET MASALAH AKU BUNUH DIRI AJA AKU UDAH GK KUAT NAMA AKU SEPTIAN HAKIM MINTA MAAF JIKA ADA SALAH
3. the amount of heat needed for 2 kg of ice at -5 celcius if it is heated until all ice melts is .... (c ice = 2.100 J/kg celcius L=336.000j/kg)
Q= m × L Q= 2 × 336.000 Q= 672.000 Are the temperature of ice just -5°C?Q = m.c.T + m.L
= 2 kg. 2100. 5 + 2.336000
= 21000 + 672000
= 693000 J
4. the amnount of heat needed for 2 kg of ice at -5 ° c , if it is heated until all the melts is ..../° c = 2100 j / kg ° c , L = 336000 j / kg ° c
Diket= m= 2kg
L= 336000 j/kg
c=2100 j/kg°C
Dita = Qt
Jawab=
Qt= Q1 + Q2
Q1= mc∆T
Q1= 2 × 2100 × (0-(-5)
Q1= 10× 2100
Q1= 21 Kj
Q2= mL
Q2= 2 × 336000
Q2= 672 Kj
Qt= 672 Kj + 21 Kj
Qt= 693 Kj
5. the amount of the heat needed for 2 kg of ice at -5 °c if it is heated until all the ice melts is?
Dik : Massa : 2kg
ΔT = 0 -(-5) = 5°C
Ces = 2100 (satuannya lupa)
L = 336.000
Q1 = m.Ces.ΔT
= 2 × 2100 × 5
= 21000 J
Q2 = m × L
= 2 × 336.000
= 672.000 J
Qtot = Q1 + Q2
= 21.000 + 672.000
= 693.000 J
Sekian.
6. Seorang anak sedang bermain skater dengan kecepatan 4 m/s jika massa skater 50 kg. berapa besarkah energi kinetik yg terbentuk
Diketahui:
m = 50 kg
v = 4 m/s
[tex] \\ [/tex]
Ditanya:
Eₖ = ...?
[tex] \\ [/tex]
Jawab:
[tex]\sf{\bold{\small{E_{k} = \frac{1}{2} × mv^{2}}}}[/tex]
[tex]\sf{\bold{\small{E_{k} = \frac{1}{2} × 50 × 4^{2}}}}[/tex]
[tex]\sf{\bold{\small{E_{k} = 400 \: Joule}}}[/tex] [tex] \\ [/tex]
Jadi, energi kinetik anak tersebut adalah 400Joule.
7. jelaskan apa yang dimaksud dengan rest,ice,compressibg,dan elevation pada penangganan cedera
Jawaban:
REST
Rest artinya mengistirahatkan bagian tubuh yang cedera, sedangkan bagian tubuh yang tidak cedera boleh tetap melakukan aktivitas
ICE
Secara umum manfaat penggunaan es pada cedera jaringan lunak adalah:
-Membatasi pembengkakan
-Mengurangi nyeri
-Mengurangi spasme otot
Compression
Kompresi adalah aplikasi gaya tekan terhadap lokasi cedera. Kompresi digunakan untuk membantu aplikasi es dan membatasi pembengkakan yang merupakan faktor utama untuk mempercepat masa rehabilitasi
Elevation
Elevasi adalah meninggikan bagian yang mengalami cedera melebihi ketinggian jantung sehingga dapat membantu mendorong cairan keluar dari daerah pembengkakan.
Penjelasan:
Semoga menjawab like ny tolong dan jawaban terbaik ya
8. what is completely filled while being and empty when at rest
class mungkin. kalo belajar penuh kalo istirahat kososng ¿
9. how much heat is required for ice at -5⁰C to water at 0⁰C?
Jawaban:
5°
because if you want to change the temperature to another temperature, use the formula for the intended temperature minus the original temperature
karena jika ingin merubah suhu ke suhu lainnya memakai rumus suhu yang dituju dikurang suhu aslinya
Penjelasan:
2,535,576°C
semoga bermanfaat
10. when an object is at rest,how will it begin to move
when the object is given a force
Jawaban:
An object will move when it is stricken/ hit by a force.
#maaf kalau salah <3
11. Ita is sick now...she Take a rest at home
Jawaban:
should
Penjelasan:
maaf klo salah
jdkan jawaban terbaik
12. aat rest is theof it?TPorad12Calenusti
Penjelasan:
Sebuah
saat istirahat adalah
dari itu?
T
Porad
12
Calenus
ti
13. dalam permainan sepatu es (ice skater),laki laki 100 kg dan perempuan 60 kg mula mula diam saling berhadapan pada suatu lantai es beku. sepuluh detik setelah keduanya saling mrndorong,mereka terpisah sejauh 80 m. berapa jauh perempuan otu bergerak dalam waktu itu?
diketahui
massa laki laki = 100kg
massa perempuan = 60kg
setelah pendorongan terpisah 80m
ditanya
berapa jauh perempuan itu bergerak ???
jawab
momentum awal = momentum akhir
(ml + mp)v = mp x vp + ml x vl
ml= massa laki-laki
mp= massa perempuan
v = kecepatan awal
vp = kecepatan perempuan setelah saling dorong
vl = kecepatan laki-laki setelah setelah saling dorong
kecepatan awal (v) = 0 (karena mereka diam dan saling berhadapan)
sehingga
(ml + mp)v = mp x vp + ml x vl
(100 + 60) 0 = 60vp + 100vl
0 = 60vp + 100vl
-60vp = 100vl
vp : vl = -100 : 60 = -5 : 3
tanda minus (-) artinya kedua orang tersebut (laki-laki dan perempuan) terdorong dengan arah yang berlawanan
jauh perempuan setelah saling dorong= 80 x (5 / (5+3))
= 80 x 5/8
= 50m
artinya jarak perempuan setelah saling dorong lebih jauh dibandingkan dengan jarak laki-laki, karena massa perempuan lebih ringan
ini ada gambar bersumber dari buku "the manga guide to physic"
smoga membantu
14. A. do you agree that he is a skater boy? B, No,I dont agree with you he is...
Jawaban:
Jawaban: biker
Penjelasan:
A. apakah kmu setuju dia skateboarder?
B. tidak aku tidak setuju dia pesepeda
15. Dua pemain sepatu es (ice skater),laki laki bermassa 100kg
Jawaban:
but uh poinhdehhejrhr
16. cari pengertian rise (rest,ice,compression dan elevation
rise:naik
rest:beristirahat
ice:es
compression:kompresi
evalution:evaluasi
17. Demain buys an ice cream. it is like a cone. it is ... ice cream
Jawaban:
It is an ice cream
Penjelasan:
Jawaban:
Demain buys an ice cream, it is like a cone. it is an ice cream.
18. At your house room/ place for take a rest is?
Jawaban:
bed room
Penjelasan:
maaf klo salah
19. Sebuah skater skater seberat 65 kg tanpa usaha selama 75 m sampai dia berhenti. Jika koefisien gesekan kinetik antara sepatu roda dan esnya adalah μk = 0,10 seberapa cepat dia bergerak di awal pantainya
m = 65 kg
s = 75 m
μ = 0,1
g = 10 m/s²
Vt =_0
Vo = __?
Gaya normal
N = m.g = 65(10) = 650 N
Gaya geseknya
f = μ.N = (0,1)(650) = 65 Newton
Usaha
W = 1/2m(Vt² - Vo²) = f.s
1/2(65)(0-Vo²) = 65(75)
Vo² = 150
Vo = √150 = 5√6 m/s
20. a block of ice which has 20 kg of mass is receiving 67 kJ of heat. The number of ice melts is... (the melting point of ice is 335 kJ/kg)
Q = m.L
67 = m.335
m = 67:335
m = 0,2 kg
Tidak ada komentar:
Posting Komentar