Home
Class 11
PHYSICS
When a rubber-band is stretched by a dis...

When a rubber-band is stretched by a distance x, it exerts a restoring force of magnitude `F = ax + bx^2` where a and b are constants. The work done in stretching the unstretched rubber band by L is :

A

`a l^(2) + bl^(3)`

B

`(1)/(2) (aL^(2)+bL^(3))`

C

`(aL^(2))/(2)+ (bL^(3))/(3)`

D

`(1)/(2) ((aL^(2))/(2) + (bL^(3))/(3))`

Text Solution

Verified by Experts

The correct Answer is:
C
`W= int_(0)^(L) (ax+ bx^(2)) dx`
`= [(ax^(2))/(2) + (bx^(3))/(3)]_(0)^(L) = (aL^(2))/(2) + (bL^(3))/(3)`
Promotional Banner

Similar Questions

Explore conceptually related problems

When a rubber bandis streched by a distance x , if exerts resuring foprce of magnitube F = ax + bx^(2) where a and b are constant . The work in streached the unstreched rubber - band by L is

Let f(x) =ax^(2) -b|x| , where a and b are constants. Then at x = 0, f (x) is

The relation between time t and distance x is t = ax^(2)+ bx where a and b are constants. The acceleration is

Let f(x)=ax^(2)-b|x| , where a and b are constant . Then at x=0 , f(x) has

A body is subjected to a conservative force given by F=b-2ax where a and b are constant, then

When a spring a stretched through a distance x, it exerts a force given by F=(-5x-16x^(3))N . What is the work done, when the spring is stretched from 0.1 m to 0.2 m?

A body covers a distance of 4m under the action of force F = (17 – 4x)N where x is the distance in metres. The work done by the force is

A force F=a+bx acts on a particle in x-direction, where a and b are constants. Find the work done by this force during the displacement from x_1 to x_2 .

A body covers a distance of 4m under the action of force F=(17-4x)N where x is the distance in metres.The work done by the force is