Amandla angaba khona entwasahlobo enwebekayo - izinkinga nezixazululo

1. Isisindo esingamakhilogremu angu-2 sinamathiselwe esiphethwini. Uma ukunwebeka kwentwasahlobo kungu-4 cm, thola amandla angaba khona esiphethwini esinwebekayo . Ukusheshisa ngenxa yamandla adonsela phansi kungu-10 m/s ².

Kwaziwa:

Isisindo (m) = 2 kg

Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 m/s ²

Isisindo (w) = mg = (2)(10) = 20 N

Ubude (x) = 4 cm = 0.04 m

Okufunwayo: Amandla angaba khona entwasahlobo enwebekayo

Isixazululo:

Ifomula yomthetho kaHooke :

F = kx

F = amandla, k = ukungaguquguquki kwentwasahlobo, x = ushintsho lobude bentwasahlobo

I-constant yasentwasahlobo:

k = w / x = 20 / 0.04 = 500 N/m

Amandla angaba khona entwasahlobo enwebekayo:

I-PE = ½ kx ² = ½ (500) (0.04) ² = (250) (0.0016) = 0.4 Joules

Isixazululo esihlukile:

I-PE = ½ kx ² = ½ (w / x) x ² = ½ wx = ½ mgx

w = isisindo, m = isisindo, x = ushintsho lobude bentwasahlobo

I-PE = ½ (2) (10) (0.04) = (10) (0.04) = 0.4 Joules.

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2. Ukwelulwa kwentwasahlobo okwelulwe ngamandla u-F = 50 N kungu-10 cm. Yimaphi amandla angaba khona entwasahlobo enwebekayo uma ukwelulwa kwentwasahlobo kungu-12 cm.

Kwaziwa:

Ubude (x) = 10 cm = 0.1 m

Amandla (F) = 50 N

Okufunwayo : Amandla angaba khona entwasahlobo enwebekayo

Isixazululo:

I-constant yasentwasahlobo:

k = F / x = 50 / 0.1 = 500 N/m

Amandla angaba khona entwasahlobo enwebekayo uma ukunwebeka kwentwasahlobo kungu-12 cm. :

I-PE = ½ kx ² = ½ (500) (0.12) ² = (250) (0.0144) = 3.6 Joules.

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3. Igrafu ka-F vs x eboniswe esithombeni esingezansi. Iyini amandla angaba khona entwasahlobo enwebekayo, uma ukunwebeka kwentwasahlobo kungu-10 cm.

Amandla angaba khona entwasahlobo enwebekayo - izinkinga nezixazululo 1

Kwaziwa:

Amandla (F) = 5 N

Ukwandiswa kwentwasahlobo (x) = 2 cm = 0.02 m

Okufunwayo: Amandla angaba khona entwasahlobo enwebekayo uma ukunwebeka kwentwasahlobo kungu-0.1 m.

Isixazululo:

I-constant yasentwasahlobo:

k = F / x = 5 / 0.02 = 250 N/m

Amandla angaba khona entwasahlobo enwebekayo uma ukunwebeka kwentwasahlobo kungu-0.1 m:

I-PE = ½ kx ² = ½ (250) (0.1) ² = (125) (0.01) = 1.25 Joules.

4. Umsubathi ugxumela ensimbini yesipilingi enesisindo esingu-500 N, intwasahlobo ifinyeza ngo-4 cm. Thola inani lamandla angase asetshenziswe ukuphoqa umsubathi.

A. 20 Joule

B. 10 Joule

C. 5 Joule

D. 2 Joule

Kwaziwa:

Amandla (F) = 500 N

Ushintsho lobude bentwasahlobo (Δx) = 4 cm = 0.04 m

Okufunwayo: Amandla angaba khona entwasahlobo

Isixazululo:

Bala i-elastic constant yesipringi usebenzisa i-equation yomthetho kaHooke:

k = F / Δx = 500 N / 0.04 m = 12500 N/m

Amandla anamandla okunwebeka entwasahlobo:

PE = ½ k Δx ² = ½ (12500) (0.04) ² = (6250) (0.0016)

PE = 10 amaJoule

Impendulo efanele ngu-B.

5. Isipilingi uma silenga ngobunzima obungamagremu angu- 500, ubude baso buyanda ngo-5 cm. Thola amandla aso angaguquki kanye namandla aso angase abe khona.

Kwaziwa:

Isisindo sento (m) = 500 amagremu = (500/1000) kg = 5/10 kg = 0.5 kg

Ukusheshisa (g) = 10 m/s ²

Isisindo sento (w) = mg = (0.5 kg)(10 m/s ² ) = 5 kg m/s ² = 5 AmaNewton

Ushintsho lobude bentwasahlobo (Δx) = 5 cm = 0.05 m

Okufunwayo: amandla angaguquki entwasahlobo (k) kanye namandla angase abe khona (PE)

Isixazululo:

Okokuqala, bala i-constant yasentwasahlobo usebenzisa i-equation yomthetho kaHooke:

k = w / Δx = 5 N / 0.05 m = 100 N/m

Ukunwebeka okungenzeka kwentwasahlobo:

PE = ½ k Δx ² = ½ (100) (0.05) ² = (50) (0.0025)

PE = 0.125 amaJoule

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