3 ceistean mu cho-aontar cunbhalach an earraich
1. Tha fad 10 cm aig fuaran ann an crochadh saor. Aig a’ cheann shaor, tha cuideam 200 gram crochte gus am bi fad an earraich 11 cm. Ma tha g = 10 m/s² , dè an neart seasmhach a th’ aig an earraich?
aithnichte:
Fad tùsail an earraich (y1 ) = 10 cm = 0.10 m
Fad deireannach an earraich (y² ) = 11 cm = 0.11 m
Atharrachadh fad an earraich (Δy) = 0.11 – 0.10 = 0.01 meatairean
Tomad an luchd (m) = 200 gram = 0.2 kg
Cuideam luchd (w) = mg = (0,2)(10) = 2 Niùtan
A dhìth: Cunbhalach an earraich (k)
fuasgladh:
Foirmle cunbhalach an earraich:
F = kΔy
k = F / Δy = 2 / 0,01 = 200 / 1 = 200 Niùtan/meatair
2. Faodar fuaran a shìneadh gus am bi e air a shìneadh 10 cm le lùth comasach de 0.5 Joule. Dè an cunbhalachd a th’ aig an earrach?
aithnichte:
Fad an earraich (Δy) = 10 cm = 0.1 meatair a chur ris
Lùth comasach an earraich (EP) = 0.5 Joules
A dhìth: Cunbhalach an earraich (k)
fuasgladh:
Tha cunbhalachd an earraich air a thomhas le bhith a’ cleachdadh foirmle Lùth Comasach an Earraich:
EP = ½ k Δy²
2 EP = k Δy 2
2 (0,5) = k (0,1) 2
1 = c (0,01)
k = 1 / 0,01 = 100 / 1 = 100 Niùtan/meatair
3. Bidh fuaran air a tharraing le feachd de 100 N a’ meudachadh a fhaid le 5 cm. Obraich a-mach cunbhalachd an earraich.
aithnichte:
Feachd (F) = 100 N
Meudachadh ann am fad an earraich (Δx) = 5 cm = 0.05 m
A dhìth: Cunbhalach an earraich (k)
fuasgladh:
F = k Δx
k = F / Δx = 100 / 0,05 = 10.000 / 5 = 2000 Niùtan/meatair
