The extension in a spring when 5g wt was hung from it was 0.56cm. If Hooke's law is obeyed, What is the extension caused by a load of 20g wt?

Frenneoma

4 years ago

F1 = e1
f2 = e2
...5g = 0.56
20g = x
cross multiply
20×0.56\ 5 = 2.24cm Ans E

Mahley101

7 years ago

exactly

Olaolul

1 year ago

pls I don't understand the explanation

DominicAfam

1 year ago

Hooke's law
F=Ke

Given,
W1=5g....to kg is 0.005kg
e1=0.56cm... to m is 0.0056m

W2=20g..... to kg is 0.02kg
e2=??

For first case,
F=Ke
Since F=ma (weight =mass)
F=0.005×10
F=0.05N

Now, F=Ke
0.05=K0.0056
Divide both side by 0.0056
K=8.93

Second case,
F=ma. (weight =mass)
F=0.02×10
F=0.2

F=Ke
0.2=8.93e
Divide both side by 8.93
e=0.0224m........ to cm
= 2.24cm

Spakfire 🔥🔥

8 months ago

@frennoma
I recommend that method for JAMBITES under time pressure

12sonia

2 months ago

the explanations are not clear