A spring 20cm long is stretched to 25cm by a load of 50N.What will be its length when stretched by 100N assuming that the elastic limit is not reached?
Force = 50N, Extension = 25cm - 20cm = 5cm
so, F= 50N/5cm = 10N/cm
E=100N/10N/cm = 10cm
new length = ( original length + E)
= 20cm + 10cm
=30cm.
k = 50N/(25-20)cm = 10N/cm.
L = 20cm + (100N/10N) * 1cm = 30 cm.
K=F/e
K=50/25-20
K=10N/cm
e=F/K
e=100/10
e=10
Therefore new length =20 + 10 = 30cm
A spring of force constant 1500newton per metre is acted by a constant force of 75newton.calculate the potential enery stored in the spring.
E=25-20=5.so,f=50N/5cm=10N/cm new length =(original length +E)=20cm+10cm=30cm
E=(25cm-20cm)=5cm so,f=50N/5cm=10N/cm E=100N/10N/cm=10cm new length=(original
length+E)=20cm+10cm=30cm
Absolutely correct!!!
I love the way physics problems are solved,for i love physics,infact.
Lennard Ans: 1.875 joules approximately 1.9 joules
And : 1.875joules
And: 1.875joules
F=50N,E=25cm-20cm=5cm.so f=50 N/5cm=10N/cm.E=100N/10N/cm=10cm=new length=(original length*E)=20cm*10cm=30cm
Force=50N ,Extension=25cm-20cm =5cm
So,F=50N/5cm=10N/cm
E=100N/10N/cm=10cm
New length=[original length+E]
=20+10
=30cm
f1/e1=f2/e2
50/25-20=100/e2
50/5=100/e2
e=100*5/50
50*e2=100*5
50e2=500
50e2/50=500/50
e2=100cm
New length=Original+Extension
=20+10
=30cm
Thanx for the solution and I look forward to seeing More of it.
10cm
The expression is correct, and absoluted.
State Hooke’s law. Explain how you can verify it in the laboratory. A force 40 N stretched a wire through 3.0 cm. What force will stretch it through 5.0 cm and through what length will a force of 100 N stretch it? What assumption have you made?
Hooks law state that force is directly proportional to extension
It can also be stated in a mathmatical way as
F ( proportional sing ) e
Introducing k as the elastic constant their for F = me
Where
F= force or load
e= extension or new length
K= elastic constant
Thanks you solved my question thanks once again ❤️❤️❤️😄
F=ke
k=50/25-20
k=10N/cm
F=ke
e=100/10
e=10cm
Total length = 10+20
E=30cm
F1/e1=f2/e2 _formula
L0=25=0.25
E1=L1- L0=0.25-0.2
=0.05
F1=50N
L2=?
F2=100N
E2=L2- 0.2
F=ke
K=f/e
Back to the formula
50/0.05=100/L2-0.2
1000/100/L2-0.2 (cross multiply)
1000(L2-0.2)=100*1
1000L2 -200=100
1000L2=100+200
1000L2=300
(Divide both side by 1000)
L2=0.3m
0.05m
Thanks
u did not solve my question
Superb
Can you please send further questions under Hooke's law and elasticity
I apologize, but your answer is incorrect. Here's the correct solution:
We are given the force constant k = 50 N / (25 cm - 20 cm) = 10 N/cm.
To find the new length when stretched by 100 N, we can use the equation e = F / k, where e is the extension.
e = 100 N / 10 N/cm = 10 cm.
Therefore, the total length of the spring is the original length plus the extension:
L = 20 cm + 10 cm = 30 cm.
So the new length is 30 cm when stretched by 100 N.
K=50N/25-20cm
=10N/cm
L=20m+100N+1cm=30cm
Done!
I don't understand what you Guy's are saying please break it to SS1 level please
Please i don't understand can you break it down to sss1 level.
F=ke
k=f/e
k=50/25-20
k=5ncm^-1
e=f/k
e=50/5
e=10
New length +original length
20+10
=30cm
A spring 20cm long is stretched to 25cm by a load of 50N. what
will be its length when stretched by 100N assuming that the elastic limit is not reached? k = F 1 / x 1 = 50 / 25 = 2 N/cm
A spring 20cm long is stretched to 25cm by a load of 50N . what will be its length when stretched by 100N assuming that the elastic limit is not reached
f=ke
k=f/e
50/25=2N/M
f=ke
100=2xe2
e2=100/2=50N/M
L2=L1+e2
20+50=70M