A spring when compressed by $4 c m$ has $2 J$ energy stored in it. The force required to extend it by $8 c m$ will be:

20 N

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 N

No worries! We‘ve got your back. Try BYJU‘S free classes today!

200 N

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2000 N

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $200 N$
Energy of spring $= \frac{1}{2} k x^{2}$

$2 = \frac{1}{2} \times k (4)^{2} \times 10^{- 4}$

$\Rightarrow \frac{1}{4 \times 10^{- 4}} = k$

Force $= k x$ $= \frac{1}{4} \times 8 \times 10^{2} = 200 N$

Suggest Corrections

Similar questions

Q. An automobile spring extends 0.2 m for 5000N load. The ratio of potential energy stored in this spring when it has been compressed by 0.2 m to the potential energy stored in the

10 μ F

capacitor at a potential difference of 10000V will be:

An automobile spring extends 0.2 m for 5000 N load. The ratio of potential energy stored in this spring when it has been compressed by 0.2 m to the potential energy stored in a 10 F capacitor at a potential difference of 10000 V will be:

Q. A spring with spring constant

k

when compresses by

1 c m

, the potential energy stored is

U

If it is further compressed by

3 c m

then change in its potential energy is

Q. A spring with spring constants K when compressed by 1 cm, the potential energy stored is U.If it is further compressed by 3 cm, then its final potential energy is

Q. A force of

10 N

holds an ideal spring with a

20 N / m

spring constant in compression, The potential energy stored in the spring is :

Join BYJU'S Learning Program

A spring when compressed by 4cm has 2J energy stored in it. The force required to extend it by 8cm will be:

The correct option is D 200NEnergy of spring =12kx22=12×k(4)2×10−4⇒14×10−4=kForce =kx =14×8×102=200 N

A spring when compressed by $4 c m$ has $2 J$ energy stored in it. The force required to extend it by $8 c m$ will be:

The correct option is D $200 N$
Energy of spring $= \frac{1}{2} k x^{2}$

$2 = \frac{1}{2} \times k (4)^{2} \times 10^{- 4}$

$\Rightarrow \frac{1}{4 \times 10^{- 4}} = k$

Force $= k x$ $= \frac{1}{4} \times 8 \times 10^{2} = 200 N$