Misalan tambayoyi kan dokar kiyaye makamashin inji

Misalan Tambayoyi 16 kan Dokar Kare Makamashin Inji

Dokar kiyaye makamashin injiniya a cikin motsi na faɗuwa kyauta

1. Kwakwa mai nauyin kilogiram 2 tana faɗuwa cikin sauƙi daga tsayin mita 6. Ƙarfin motsi na kwakwa idan ta isa ƙasa shine… g = 10 m/s²
Tattaunawa
An sani :
Nauyin kwakwa (m) = 2 kg
Tsawo (h) = mita 6
Saurin gudu saboda nauyi (g) = 10 m/s²
An tambaya : kuzarin motsi (EK)
Jawab :

Makamashin injina na farko
Idan abu yana cikin faɗuwar 'free fall', ba shi da saurin farko. Domin babu saurin farko, babu kuzarin motsi na farko. Da farko, abu yana kan tsayin mita 6, don haka ba shakka, abu yana da ƙarfin ƙarfin nauyi na farko. Don haka, makamashin injiniya na farko (EM) shine_o) shine makamashin ƙarfin nauyi (EP).
EM_o = EP = mgh = (2)(10)(6) = Joules 120

Makamashin injiniya na ƙarshe
Lokacin da ake tafiya ƙasa, tsayin abu yana raguwa, don haka ƙarfin ƙarfinsa na nauyi yana raguwa. Ƙarfin ƙarfin nauyi ba ya ɓacewa amma yana canzawa zuwa kuzarin motsi. Lokacin da ya isa ƙasa, tsayin abu iri ɗaya ne da tsayin ƙasa (h = 0), don haka ƙarfin ƙarfinsa na nauyi sifili ne (h = 0, don haka EP = mgh = 0). Duk ƙarfin ƙarfin nauyi ana canza shi zuwa makamashin motsi. Don haka, ƙarfin injiniya na ƙarshe (EM) shine_t) shine makamashin motsi (EK).
EM_t = EK = ½ mv²

Dokar kiyaye makamashin inji :
Makamashin injina na farko = makamashin injina na ƙarshe
EM_o = EM_t
EP = EK
120 = EK
Ƙarfin motsi da kwakwa ke da shi idan ya isa ƙasa shine Joules 100.

2. Ƙwallo mai nauyin kilogiram 2 faɗuwar rana kyauta daga matsayi na A kamar yadda aka nuna a hoton. (g = 10 ms^-2) Lokacin da aka isa wurin B, kuzarin motsi yana daidai da sau 2 na ƙarfin da ake da shi, don haka tsayin wurin B daga ƙasa shine…

A. 80 m
B. 70 m
C. mita 60
D. 40 m
E. 30 m

Tattaunawa
An sani :
Nauyin ƙwallon (m) = 2 kg
Saurin gudu saboda nauyi (g) = 10 ms^-2
Tsawo A (h)_A) = mita 90
An tambaya : Tsawo B (h)_B)
Jawab :

Ketika bola tiba di titik B, energi kinetik bola di titik B sama dengan 2 kali energi potensial gravitasi bola di titik B.
EK = 2 EP
½ mv² = 2 mgh_B
½ v² = 2 gh_B
v² = 2(2)(10) h_B
v² = awanni 40_B
Hitung gudun (v) ƙwallon lokacin da ta isa wurin B bayan faɗuwa daga wurin A.
v² = 2 gh = 2(10)(90–h_B) = 20(90–awa_B)
Sauya v² a cikin lissafin da ke sama da v² a cikin wannan lissafin.
v² = awanni 40_B
20(90–h)_B) = awanni 40_B
1800-20 h_B = awanni 40_B
1800 = awanni 40_B + 20 h_B
1800 = awanni 60_B
h_B = 1800/60
h_B = 30 mita
Amsar da ta dace ita ce E.

3. Abu mai nauyin kilogiram 1 yana faɗuwa cikin sauƙi daga saman ginin bene mai hawa da yawa wanda tsayinsa ya kai mita 80. Idan aka yi watsi da gogayya da iska kuma saurin gudu saboda nauyi (g) ya kai mita 10/s² Don haka kuzarin motsi na abu idan ya isa saman ƙasa shine...
Tattaunawa
An sani :
Mass (m) = 1 kg
Tsawo (h) = mita 80
Saurin gudu saboda nauyi (g) = 10 m/s²
An tambaya : kuzarin motsi (EK) na abu lokacin da abin ya isa saman ƙasa
Jawab :
Abubuwa suna faɗuwa cikin 'yanci don haka abu ba shi da saurin farko. Don haka, idan yana da tsayin mita 80, abu yana da ƙarfin ƙarfin nauyi amma abu ba shi da kuzarin motsi (v = 0 don haka EK = ½ mv² = 0). Don haka makamashin injiniya na farko (EM)_o) = makamashin ƙarfin nauyi (EP).

EM _o = EP = mgh = (1)(10)(80) = Joules 800

A lokacin faɗuwar 'yanci, makamashin ƙarfin nauyi yana canzawa zuwa makamashin motsi. Idan wani abu ya bugi ƙasa, duk ƙarfin ƙarfin nauyi yana canzawa zuwa makamashin motsi. Saboda haka, idan ya isa ƙasa, abu yana da kuzarin motsi amma babu kuzarin ƙarfin nauyi (h = 0, don haka EP = mgh = 0). Saboda haka, makamashin injiniya na ƙarshe (EM _t ) = makamashin motsi (EK)

Dokar kiyaye makamashin injiniya ta bayyana cewa makamashin injiniya na farko daidai yake da makamashin motsi na ƙarshe.
EM_o = EM_t
EP = EK
800 = EK
Ƙarfin motsi (EK) na abu idan ya isa ƙasa shine Joules 800.

4. Abu yana faɗuwa cikin sauƙi daga tsayin mita 10 sama da ƙasa. Idan nauyin abin ya kai kilogiram 4 kuma saurin gudu saboda nauyi (g) = 10 ms ^–2 , to kuzarin motsi da saurin abin a tsayin mita 5 sama da ƙasa…

Tattaunawa
An sani :
Canjin tsayi (h) = 10 – 5 = mita 5
Mass (m) = 4 kg
Saurin gudu saboda nauyi (g) = 10 m/s²
An tambaya : kuzarin motsi na abu a tsayin mita 5
Jawab :
(a) Ƙarfin motsi na abu a tsayin mita 5
Makamashin injina na farko (EM)_o) = makamashin ƙarfin nauyi (EP)
EM_o = EP = mgh = (4)(10)(5) = Joules 200
Makamashin injiniya na ƙarshe (EM)_t) = kuzarin motsi (EK)
EM_t = EK
Dokar kiyaye makamashin injiniya ta bayyana cewa makamashin injiniya na farko iri ɗaya ne da makamashin injiniya na ƙarshe.
EM_o = EM_t
200 = EK
Ƙarfin motsi na abu a tsayin mita 5 sama da ƙasa shine Joules 200.

(b) Saurin abin a tsayin mita 5
Makamashin injina na farko (EM)_o) = makamashin injiniya na ƙarshe (EM)_t)
EP = EK
200 = ½ mv²
2(200) / 4 = v²
100 = ku²
v = √100
v = mita 10/daƙiƙa
Gudun abin shine mita 10/daƙiƙa

5. 'Ya'yan mangwaro suna faɗuwa cikin sauƙi daga tsayin mita 2. Idan g = 10 ms^-2, ƙididdige saurin mangwaro kafin ya faɗi ƙasa ta amfani da dokar makamashin inji.
Tattaunawa
An sani :
Tsawo (h) = mita 2
Saurin gudu saboda nauyi (g) = 10 m/s²
An tambaya : saurin mangwaro kafin ya faɗi ƙasa (v)
Jawab :
Makamashin injina na farko (EM)_o) = makamashin ƙarfin nauyi (EP)
EM_o = EP = mgh = m (10)(2) = 20 m
Makamashin injiniya na ƙarshe (EM)_t) = kuzarin motsi (EK)
EM_t = EK = ½ mv²
Dokar kiyaye makamashin injiniya ta bayyana cewa makamashin injiniya na farko iri ɗaya ne da makamashin injiniya na ƙarshe.
EM_o = EM_t
20 m = ½ mv²
20 = ½ v²
2(20) = v²
40 = ku²
v = √40 = √(4)(10) = mita 2√10/daƙiƙa
Gudun mangwaron kafin ya faɗi ƙasa shine mita 2√10/daƙiƙa.
Yana da mahimmanci a lura cewa a cikin lissafinmu, muna yin watsi da gogayya ta iska. A zahiri, idan wani abu ya faɗi cikin 'yanci a saman Duniya, gogayya ta iska kuma tana shafar motsinsa, wanda ke haifar da saurin da ya yi ƙasa da lissafin da ke sama.

Dokar kiyaye makamashin injiniya a motsi a tsaye sama da ƙasa

6. Ana jefa wani abu mai nauyin kilogiram 1 a tsaye ƙasa da sauri na farko na m2/s daga gefen ginin mai tsayin mita 40. Ƙarfin motsi na abin idan yana a tsayin mita 10 sama da ƙasa shine… g = 10 m/s ²

Tattaunawa
An sani :
Nauyin abu (m) = 1 kg
Saurin farko (v)_o) = 2 m/s
Canjin tsayi = 40 – 10 = mita 30
Saurin gudu saboda nauyi (g) = 10 m/s²
An tambaya : kuzarin motsi na abu a tsayin mita 10 sama da matakin ƙasa
Jawab :

Makamashin injina na farko
Da farko, abu yana da kuzarin ƙarfin nauyi da kuma kuzarin motsi.
Makamashin ƙarfin nauyi na farko (EP) = mgh = (1)(10)(30) = Joules 300
Makamashin motsi na farko (EK) = ½ mv_o² = ½ (1)(2)² = ½ (4) = Joule 2
Don haka makamashin injiniya na farko = ƙarfin ƙarfin nauyi na farko + kuzarin motsi na farko = 300 + 2 = Joules 302.

Makamashin injiniya na ƙarshe
Yayin da abu ke tafiya ƙasa, tsayinsa yana raguwa, don haka yana rage ƙarfin ƙarfinsa na nauyi. Wannan ƙarfin ƙarfin nauyi yana canzawa zuwa kuzarin motsi. Bayan motsi mita 30, ko kuma lokacin da abu ya kai mita 10 sama da ƙasa, duk ƙarfin ƙarfinsa na nauyi yana canzawa zuwa kuzarin motsi.
Don haka makamashin injiniya na ƙarshe = makamashin motsi na ƙarshe = ƙarfin ƙarfin nauyi na farko + kuzarin motsi na farko = 300 + 2 = Joules 302.

7. Ana jefa wani abu mai nauyin kilogiram 1 a tsaye sama da saurin farko na m10/s. A tantance (a) ƙarfin ƙarfin nauyi a mafi girman matsayi (b) mafi girman matsayi ko tsayin da abin ya kai… g = 10 m/s²
Tattaunawa
An sani :
Nauyin abu (m) = 1 kg
Saurin farko (v)_o) = 10 m/s
Saurin gudu saboda nauyi (g) = 10 m/s²
An tambaya : kuzarin ƙarfin nauyi a mafi girman matsayi da kuma mafi ƙasƙanci
Jawab :
(a) Makamashin ƙarfin jijiya a mafi girman matsayi
Makamashin injina na farko
Da farko abu yana da kuzarin motsi amma abu ba shi da kuzarin ƙarfin nauyi. Kuzarin injiniya na farko = kuzarin motsi na farko (EK) = ½ mv_o² = ½ (1)(10)² = ½ (100) = Joules 50.
 
Makamashin injiniya na ƙarshe
Bayan ya tashi tsaye sama, tsayin abu yana ƙaruwa, ta haka yana ƙara ƙarfin ƙarfin nauyi. Wannan ƙarfin ƙarfin nauyi ba ya tasowa kwatsam amma yana fitowa ne daga kuzarin motsi. A matsakaicin tsayinsa ko mafi girman matsayi, abu yana hutawa na ɗan lokaci kafin ya juya alkibla. Domin yana hutawa na ɗan lokaci, saurinsa sifili ne, kuma kuzarin motsi shi ma sifili ne. Duk kuzarin motsi ana mayar da shi zuwa kuzarin ƙarfin nauyi.
Makamashin injiniya na ƙarshe = makamashin ƙarfin nauyi (EP)

Dokar kiyaye makamashin inji:
Makamashin injina na farko = makamashin injina na ƙarshe
EK = EP
50 = EP
Ƙarfin ƙarfin nauyi shine joules 50.

(b) Mafi girman matsayi ko matsakaicin tsayi
EP = mgh
50 = (1)(10) h
50 = awanni 10
h = 50 / 10 = mita 5
Mafi girman matsayi ko tsayin da abu ya kai shine mita 5 daga saman ƙasa.

 

Dokar kiyaye makamashin injiniya a motsi a kan jirgin sama mai karkata

8. Daga yanayin hutawa, toshe yana zamewa ba tare da wata gogayya ba a kan jirgin da ya karkata. Saurin toshewar idan ya isa ƙasan jirgin da ya karkata shine… g = 10 m/s²
Tattaunawa
An sani :
Tsawo (h) = mita 8
Saurin gudu saboda nauyi (g) = 10 m/s²
An tambaya : gudun (v) na toshe lokacin da ya kai ƙasan jirgin da aka karkata
Jawab :
Idan tsayinsa ya kai mita 8, toshen yana da ƙarfin ƙarfin nauyi amma babu kuzarin motsi saboda bai motsa ba tukuna. Saboda haka, ƙarfin injiniya na farko na toshen shine ƙarfin ƙarfin nauyi.
Makamashin injina na farko (EM)_o) = makamashin ƙarfin nauyi (EP)
EM_o = EP = mgh = m (10)(8) = 80 m
Yayin da tubalin ya fara motsawa ƙasa, ƙarfin ƙarfinsa na nauyi ya fara canzawa zuwa kuzarin motsi. Lokacin da ya kai ƙasan karkata, duk ƙarfin ƙarfinsa na nauyi zai koma makamashin motsi. Don haka, ƙarfin injiniya na ƙarshe na tubalin shine makamashin motsi.
Makamashin injiniya na ƙarshe (EM)_t) = kuzarin motsi (EK)
EM_t = EK = ½ mv²
Dokar kiyaye makamashin injiniya ta bayyana cewa makamashin injiniya na farko iri ɗaya ne da makamashin injiniya na ƙarshe.
EM_o = EM_t
80 m = ½ mv²
80 = ½ v²
160 = ku²
v = √160 = √(16)(10) = 4√10 m/s
Gudun tubalan da ke ƙasan jirgin sama mai karkata shine mita 4√10/daƙiƙa.

9. Wani abu mai nauyin kilogiram 1, wanda da farko yana hutawa, ana fitar da shi daga saman wani jirgin sama mai santsi wanda tsawonsa ya kai mita 8. A tantance kuzarin motsi bayan abin ya zame a kan jirgin sama mai karkata na tsawon mita 5…. (g = 10 m/s²)
Tattaunawa
An sani :
Nauyin abu (m) = 0,2 kg
Tsawon hypotenuse da abin ya ratsa = mita 5
Saurin gudu saboda nauyi (g) = 10 m/s²
An tambaya : kuzarin motsi (EK) na abu bayan zamewa mita 5
Jawab :
Mataki na farko wajen magance wannan matsala shine a ƙididdige canjin tsayin abin bayan abin ya zame a kan wani jirgin sama mai karkata na tsawon mita 5.
ba 30^o = tsawon gefen tsaye / tsawon hypotenuse
0,5 = tsawon gefen tsaye / 5
Tsawon gefen tsaye = (0,5)(5) = mita 2,5
Canjin tsayin abu shine mita 2,5.

Makamashin injina na farko (EM)_o) = makamashin ƙarfin nauyi (EP)
EM_o = EP = mgh = (1)(10)(2,5) = Joules 25
Makamashin injiniya na ƙarshe (EM)_t) = kuzarin motsi (EK)
EM_t = EK
Dokar kiyaye makamashin injiniya ta bayyana cewa makamashin injiniya na farko iri ɗaya ne da makamashin injiniya na ƙarshe.
EM_o = EM_t
25 = EK
Ƙarfin motsi na abu bayan abin ya zame mita 5 shine Joules 25.

 

Dokar kiyaye makamashin injiniya a motsi a cikin jirgin sama mai lanƙwasa

10. Abu mai nauyin kilogiram 1 yana zamewa ba tare da gogayya daga tsayin mita 5 ba tare da wani saurin farko ba. Ƙarfin motsi da saurin abu a mafi ƙasƙanci sune… g = 10 m/s²
Tattaunawa
An sani :
Nauyin abu (m) = 1 kg
Canjin tsayi (h) = mita 5
Saurin gudu saboda nauyi (g) = 10 m/s²
An tambaya : kuzarin motsi (EK) da saurin (v) na abu a mafi ƙasƙanci wuri
Jawab :

(a) Ƙarfin motsi na abu a mafi ƙasƙanci wuri
Idan yana da tsayin mita 5 kuma bai fara motsi ba tukuna, abu yana da ƙarfin ƙarfin nauyi amma babu ƙarfin motsi saboda har yanzu yana hutawa. Don haka, ƙarfin injiniya na farko kafin abu ya fara motsi shine ƙarfin ƙarfin nauyi. Da zarar abu ya fara motsawa ƙasa, ƙarfin ƙarfin nauyi zai fara canzawa zuwa makamashin motsi. Lokacin da ya kai mafi ƙasƙanci, duk ƙarfin ƙarfin nauyi zai canza zuwa makamashin motsi. Don haka, ƙarfin injiniya na ƙarshe shine makamashin motsi.
Makamashin injina na farko = makamashin ƙarfin nauyi
EM_o = EP = mgh = (1)(10)(5) = Joules 50
Makamashin injiniya na ƙarshe = makamashin motsi
EM_t = EK = ½ mv_t²
Dokar kiyaye makamashin injiniya ta bayyana cewa makamashin injiniya na farko iri ɗaya ne da makamashin injiniya na ƙarshe..
EM_o = EM_t
EP = EK
50 = EK
Ƙarfin motsi (EK) a mafi ƙanƙanta shine Joules 50.

(b) Saurin abu a mafi ƙasƙanci wuri
Hanya ta farko:
Dokar kiyaye makamashin inji:
Makamashin injina na farko (EM)_o) = makamashin injiniya na ƙarshe (EM)_t)
Makamashin ƙarfin nauyi (EP) = makamashin motsi (EK)
50 = ½ mv²
2(50) / m = v²
100 / 1 = v²
100 = ku²
v = √100
v = mita 10/daƙiƙa
 
Hanya ta biyu:
Dokar kiyaye makamashin inji :
Makamashin injina na farko (EM)_o) = makamashin injiniya na ƙarshe (EM)_t)
Makamashin ƙarfin nauyi (EP) = makamashin motsi (EK)
mgh = ½ mv²
gh = ½ v²
2 gh = v²
2(10)(5) = v²
100 = ku²
v = √100
v = mita 10/daƙiƙa
Saurin abu a mafi ƙasƙanci shine 10 m/s.

11. Abu mai nauyin kilogiram 2 wanda da farko yana hutawa, yana zamewa ba tare da gogayya ba daga tsayin mita 10 sama da ƙasa. Ƙarfin motsi da saurin abin a tsayin mita 2 sama da ƙasa shine… g = 10 m/s²
Tattaunawa
An sani :
Nauyin abu (m) = 2 kg
Canjin tsayi (h) = 10 – 2 = mita 8
Saurin gudu saboda nauyi (g) = 10 m/s²
An tambaya : kuzarin motsi (EK) da saurin (v) na abu a tsayin mita 2 sama da saman ƙasa.
Jawab :
Maganin wannan matsala yayi kama da maganin matsalar da ta gabata.

(a) Ƙarfin motsi na abu a tsayin mita 2
Makamashin injina na farko = makamashin ƙarfin nauyi
EM_o = EP = mgh = (2)(10)(8) = Joules 160
Makamashin injiniya na ƙarshe = makamashin motsi
EM_t = EK = ½ mv_t²
Dokar kiyaye makamashin injiniya ta bayyana cewa makamashin injiniya na farko iri ɗaya ne da makamashin injiniya na ƙarshe..
EM_o = EM_t
EP = EK
160 = EK
Makamashin motsi (EK) a tsayin mita 2 shine Joules 160.

(b) Saurin abu a mafi ƙasƙanci wuri
Hanya ta farko:
Dokar kiyaye makamashin inji:
Makamashin injina na farko (EM)_o) = makamashin injiniya na ƙarshe (EM)_t)
Makamashin ƙarfin nauyi (EP) = makamashin motsi (EK)
160 = ½ mv²
160 = ½ (2) v²
160 = ku²
v = √160 = √(16)(10) = mita 4√10/daƙiƙa

Hanya ta biyu:
Dokar kiyaye makamashin inji :
Makamashin injina na farko (EM)_o) = makamashin injiniya na ƙarshe (EM)_t)
Makamashin ƙarfin nauyi (EP) = makamashin motsi (EK)
mgh = ½ mv²
gh = ½ v²
2 gh = v²
2(10)(8) = v²
160 = ku²
v = √160 = √(16)(10) = mita 4√10/daƙiƙa
Gudun abu a tsayin mita 2 shine mita 4√10/daƙiƙa.

12. A lokacin hunturu, a Sweden, ana gudanar da gasannin tsere kan kankara a tsaunuka. Masu tsere kan kankara suna zamewa daga tsayin A kamar yadda aka nuna a hoto na 2.

Idan saurin farko na mai tsere kan dusar ƙanƙara = sifili, kuma saurin da aka samu sakamakon nauyi = 10 ms ^-2 , to saurin mai tsere kan dusar ƙanƙara a tsayin B shine…

A. √2 ms ^-1

B. 5 √2 ms ^-1

C. 10 √2 ms ^-1

D. 20 √2 ms ^-1

E. 25 √2 ms ^-1

Tattaunawa

An san cewa:

Saurin farko (v _o ) = 0

Tsawon farko (h _o ) = 50 m – 10 m = 40 m

Tsawon ƙarshe (h _t ) = 0

Saurin gudu saboda nauyi (g) = 10 ms ^-2

Tambaya: Saurin ƙarshe (v _t )

Amsa:

Canji a cikin kuzarin motsi:

ΔEK = 1/2 m (v_t² - v_o²) = 1/2 m (v_t² - 0) = 1/2 mv_t²

Canji a cikin makamashi mai yuwuwa:

Δ EP = mg (h _t – h _o ) = m (10)(0-40) = m (10)(-40) = – m400

Lissafa saurin ƙarshe (v _t ) ta amfani da dabarar dokar kiyaye makamashin inji:

0 = Δ EK + Δ EP

0 = 1/2 mv _t ² – 400 m

1/2 mv _t ² = 400 m

1/2 v _t ² = 400

v _t ² = 2(400)

v _t = √(400)(2)

v _t = 20√2 m/s

Amsar da ta dace ita ce D.

13. Ana fitar da ƙwallon da nauyinta ya kai kilogiram 1 sannan a zame daga matsayi na A zuwa matsayi na C a kan hanya mai santsi kamar yadda aka nuna a hoton da ke ƙasa. hanzari saboda nauyi = 10 ms^-2, to kuzarin motsi (Ek) na ƙwallon idan ya kai ga matsayi na C shine….

A. 25,0 joules
B. 22,5 joules
C. 20,0 joules
D. 12,5 joules
E. 7,5 joules

Tattaunawa
An san cewa:
Nauyin ƙwallon (m) = 1 kg
Saurin gudu saboda nauyi (g) = 10 m/s²
Canjin tsayi (h) = 0,75 m

Makamashin ƙarfin nauyi na farko (EP)_o) = mgh = (1)(10)(0,75) = 7,5 Joules
Makamashin motsi na farko (EK)_o) = ½ mv_o² = ½ m (0)²) = 0
Makamashin ƙarfin nauyi na ƙarshe (EP)_t) = mgh = mg (0) = 0
Tambaya: Makamashin motsa jiki (EK) na ƙwallon lokacin da yake a wurin C
Amsa:

Dokar kiyaye makamashin injiniya ta bayyana cewa makamashin injiniya na ƙwallon a matsayinsa na farko daidai yake da makamashin injiniya na ƙarshe a matsayinsa na ƙarshe. A cikin wannan matsalar, matsayin farko shine matsayin ƙwallon kafin a sake shi, kuma matsayin ƙarshe shine maki C.

Makamashin injina na farko (EM)_o) = Makamashin injiniya na ƙarshe (EM)_t)
Makamashin ƙarfin nauyi na farko (EP)_o) + Ƙarfin motsi na farko (EK)_o) = Ƙarfin ƙarfin nauyi na ƙarshe (EP)_t) + Makamashin motsi na ƙarshe (EK)_t)
Joules 7,5 + 0 = 0 + EK_t
Joules 7,5 = EK_t
Ƙarfin motsi (EK) na ƙwallon a wurin C shine Joules 7,5.
Amsar da ta dace ita ce E.

14. An saki ƙwallon A mai nauyin kilogiram 2 kuma tana tafiya a kan hanya mai santsi kamar yadda aka nuna a hoton. Idan saurin gudu saboda nauyi shine 10 m/s^-2, kuzarin motsi na ƙwallon lokacin da a cikin B shine….

A. 4 joules
B. 8 joules
C. 10 joules
D. 12 joules
E. 24 joules

Tattaunawa
An san cewa:
Nauyin ƙwallon (m) = 2 kg
Saurin gudu saboda nauyi (g) = 10 m/s²
Canjin tsayi (h) = 120 cm = mita 1,2

Makamashin ƙarfin nauyi na farko (EP)_o) = mgh = (2)(10)(1,2) = 24 Joules
Makamashin motsi na farko (EK)_o) = ½ mv_o² = ½ m (0)²) = 0
Makamashin ƙarfin nauyi na ƙarshe (EP)_t) = mgh = mg (0) = 0
Tambaya: Ƙarfin motsi na ƙwallon a B
Amsa:
Makamashin injinan ƙwallon a wurin A (makamashin injinan farko) iri ɗaya ne da makamashin injinan ƙwallon a wurin B (makamashin injinan ƙarshe).

Makamashin injina na farko (EM)_o) = Makamashin injiniya na ƙarshe (EM)_t)
Makamashin ƙarfin nauyi na farko (EP)_o) + Ƙarfin motsi na farko (EK)_o) = Ƙarfin ƙarfin nauyi na ƙarshe (EP)_t) + Makamashin motsi na ƙarshe (EK)_t)
Joules 24 + 0 = 0 + EK_t
EK_t = Joule 24
Ƙarfin motsi (EK) na ƙwallon a wurin C shine Joules 24.
Amsar da ta dace ita ce E.

 

Dokar kiyaye makamashin injiniya a cikin motsi na parabolic

15. Ana harba ƙwallon da nauyinta ya kai kilogiram 0,1 a gudun mita 10/s. Idan kusurwar tsayinta ta kai mita 30.^o kuma saurin gudu saboda nauyi shine 10 m/s² , ƙayyade (a) ƙarfin ƙarfin nauyi na ƙwallon a mafi girman matsayi (b) mafi girman matsayi ko matsakaicin tsayi.
Tattaunawa
An sani :
Nauyin ƙwallon (m) = 0,1 kg
Saurin farko (v)_o) = 10 m/s
Kusurwar tsayi = 30^o
Saurin gudu saboda nauyi (g) = 10 m/s²
An tambaya : kuzarin motsi (EK) na ƙwallon a mafi girman matsayi
Jawab :

(a) Ƙarfin ƙarfin nauyi na ƙwallon a mafi girman matsayi
Hanyar ƙwallon tana kama da parabola kamar yadda aka nuna a hoton da ke ƙasa.
Da farko ƙididdige ɓangaren saurin farko a cikin alkiblar kwance (v)_ox) da kuma tsaye (v_oy).
Idan hanyar motsi ta parabolic ta yi kama da hoton da ke sama, to ana nazarin motsi a cikin alkiblar kwance ko kwance kamar haka. motsi na layi ɗaya kuma ana nazarin motsi a tsaye kamar yadda ake yi motsi a tsaye samaA cikin motsi mai layi ɗaya, saurin yana nan a ko'ina cikin motsi. Akasin haka, a cikin motsi mai tsayi sama, lokacin da abin ya kai matsakaicin tsayinsa ko mafi girman matsayi, yana tsayawa na ɗan lokaci kafin ya juya alkibla. Don haka, a mafi girman matsayi, saurin tsaye na abu sifili ne, amma a mafi girman matsayi, abin har yanzu yana da saurin v._ox.

Makamashin injina na farko
Lokacin da ƙwallon ya fara motsi, tsayinsa iri ɗaya ne da ƙasa ko saman ƙasa (h = 0), don haka ƙarfin ƙarfinsa na nauyi sifili ne (h = 0, don haka EP = mgh = 0). Akasin haka, lokacin da ya fara motsi, ƙwallon yana da matsakaicin saurin farko, don haka ƙarfin motsi shine matsakaicin. Saboda haka, ƙarfin motsi na farko na ƙwallon shine makamashin motsi.
Makamashin injina na farko (EM)_o) = kuzarin motsi (EK)
EM_o = EK = ½ mv_o² = ½ (0,1)(10)² = ½ (0,1)(100) = ½ (10) = Joules 5

Makamashin injiniya na ƙarshe
Bayan fara motsi, tsayin ƙwallon yana ƙaruwa, don haka ƙarfin ƙarfinsa na nauyi yana ƙaruwa. Idan ya kai matsakaicin tsayinsa, ƙarfin ƙarfinsa na nauyi yana kaiwa matsakaicin ƙimarsa. Yaya game da kuzarin motsi na ƙwallon? A mafi girman matsayi, saurin tsaye na ƙwallon (voy) sifili ne, amma har yanzu yana da saurin kwance (vox).
Makamashin motsi a matsakaicin tsayi :
EK = ½ mv_ox² = ½ (0,1)(5√3)² = ½ (0,1)((25)(3)) = ½ (0,1)(75) = Joules 3,75

Dokar kiyaye makamashin inji:
Makamashin injina na farko (EM)_o) = makamashin injiniya na ƙarshe (EM)_t)
EK = EP + EK
5 = EP + 3,75
EP = 5 – 3,75 = Joules 1,25
Ƙarfin ƙarfin nauyi a matsakaicin tsayi shine Joules 1,25.
 
(b) Mafi girman matsayi ko matsakaicin tsayi
EP = mgh
1,25 = (0,1)(10) h
1,25 = h
Tsawon mafi girma shine mita 1,25.

16. Ana jefa ƙwallon da nauyinta ya kai kilogiram 0,1 a kwance a gudun mita 10 daga gefen ginin da ke da tsawon mita 10. Idan saurin gudu saboda nauyi a wannan wurin ya kai mita 10/s², to kuzarin motsi na ƙwallon idan ya isa ƙasa shine….
Tattaunawa
An sani :
Nauyin ƙwallon (m) = 0,1 kg
Saurin farko (v)_o) = 10 m/s
Saurin gudu saboda nauyi (g) = 10 m/s²
Canjin tsayi (h) = 10 – 2 = mita 8
An tambaya : kuzarin motsi na ƙwallon a tsayin mita 2
Jawab :
Idan ƙwallon ta kai tsayin mita 10, tana da ƙarfin ƙarfin nauyi. Ƙwallon tana fara motsawa bayan an jefa ta da saurin farko na m10/s, don haka tana da ƙarfin motsi. Don haka, ƙarfin farko na ƙwallon shine ƙarfin ƙarfin nauyi + ƙarfin motsi.
Makamashin ƙarfin nauyi na farko (EP) = mgh = (0,1)(10)(10) = Joules 10
Makamashin motsi na farko (EK) = ½ mv_o² = ½ (0,1)(10)² = ½ (0,1)(100) = ½ (10) = Joules 5

Yayin da ƙwallon ke tafiya ƙasa, tsayinsa yana raguwa, don haka yana rage ƙarfin ƙarfinsa na nauyi. Idan ya isa ƙasa, tsayinsa iri ɗaya ne da tsayin ƙasa (h = 0), don haka ƙarfin ƙarfinsa na nauyi sifili ne (h = 0, don haka EP = mgh = 0). Duk ƙarfin ƙarfin nauyi ana mayar da shi zuwa makamashin motsi.
Makamashin motsi na ƙarshe = ƙarfin ƙarfin nauyi na farko + kuzarin motsi na farko = 10 + 5 = Joules 15