Amandla angenzeka - izinkinga nezixazululo
Amandla adonsela phansi anamandla
1. Amandla angu-4900 Joules asetshenziswa ukuphakamisa into enesisindo esingu-50 kg ibe ukuphakama okungu-h. Kuyini ukuphakama okungu-h? Ukusheshisa ngenxa yamandla adonsela phansi (g) = 9.8 ms -2.
Kwaziwa:
Ushintsho lwamandla angase abe khona (ΔPE) = 4900 Joules
Isisindo sento (m) = 50 kg
Ukusheshisa ngenxa yamandla adonsela phansi (g) = 9.8 m/s 2
Okufunwayo: ukuphakama (Δh)
Isixazululo:
ΔPE = mg Δh
4900 = (50)(9.8) Δh
4900 = 490 Δh
Δh = amamitha ayi-10
Amandla angase abe khona entwasahlobo
2. Igrafu engezansi ikhombisa ubudlelwano phakathi kwamandla (F) kanye no-x (ushintsho lobude) besiphethu. Uma ushintsho lobude besiphethu lungamasentimitha angu-8, iyini amandla angase abe khona esiphethu?
Kwaziwa:
Amandla (F) = 2 Newton
Ushintsho lobude 1 (x) = 1 cm = 1/100 m = 0.01 m
Ushintsho lobude 2 = 8 cm = 8/100 m = 0.08 m
Okufunwayo: Amandla entwasahlobo
Isixazululo:
Okungaguquki kwentwasahlobo:
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Amandla angase abe khona entwasahlobo:
∆PE = 1/2 kx 2
∆PE = 1/2 (200 N/m)(0.08 m) 2
∆PE = (100 N/m)(0.0064 m2 )
∆PE = 0.64 Nm
3. Ngokusekelwe kuthebula elingezansi, u-F = isisindo sento, u-∆L = ushintsho lobude bentwasahlobo. Uyini umsebenzi owenziwe entwasahlobo ukuze ushintsho lobude bentwasahlobo lube ngu-10 cm.
Kwaziwa:
Ushintsho lobude bentwasahlobo (∆L) = 10 cm = 0.1 m
Okufunwayo: Umsebenzi owenziwe entwasahlobo
Isixazululo:
Okungaguquki kwentwasahlobo:
k = F / ∆x = 20 N / 0.04 m = 500 N/m
k = F / ∆x = 30 N / 0.06 m = 500 N/m
k = F / ∆x = 40 N / 0.08 m = 500 N/m
I-constant yaseNtwasahlobo ingu-500 N/m
Umsebenzi wenziwa entwasahlobo ngakho-ke ushintsho lobude bentwasahlobo luyi-10 cm:
W = 1/2 k ∆L 2 = 1/2 (500 N/m)(0.1 m) 2 = (250 N/m)(0.01 m 2 ) = 2.5 N m = 2.5 Joules.
4. Igrafu engezansi ikhombisa ubudlelwano phakathi kwamandla (F) kanye noshintsho lobude (x). Iyini amandla entwasahlobo asekelwe kugrafu engezansi.
Kwaziwa:
F = 40 Newton
Δx = 0.08 m
Kufunwa: amandla entwasahlobo
Isixazululo:
Okungaguquki kwentwasahlobo:
k = F / Δx
Amandla angase abe khona entwasahlobo:
PE = ½ k Δx 2 = ½ (F/Δx) Δx 2 = ½ F Δx
I-PE = ½ (40)(0.08) = (20)(0.08) = 1.6 Joule
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