When a spring is stretched by a distance x, it exerts a force, given by F=(-5 x-16 x3) N The work done, when the spring is stretched from 0.1 ,m to 0.2 m is

Question

When a spring is stretched by a distance x, it exerts a force, given by F=(-5 x-16 x3) N The work done, when the spring is stretched from 0.1 ,m to 0.2 m is (A)  8.7× 10-2J  (B)  12.2× 10-2J  (C)  8.7× 10-1J  (D)  12.2× 10-1J

Tardigrade Admin · Accepted Answer

Work done is equal to difference in potential energies for two different positions of spring. Given,

F = - 5 x - 16 x^{3}

= - (5 + 16 x^{2}) x

= - k x

where

k (= 5 + 16 x^{2})

is force constant of spring. Therefore, work done in stretching the spring from position

x_{1}

to position

x_{2}

is

W = \frac{1}{2} k_{2} x_{2}^{2} - \frac{1}{2} k_{1} x_{1}^{2}

we have,

x_{1} = 0.1 m

and

x_{2} = 0.2 m

,

∴ W = \frac{1}{2} [5 + 16 (0.2)^{2}] (0.2)^{2}

- \frac{1}{2} [5 + 16 (0.1)^{2}] (0.1)^{2}

= 2.82 \times 4 \times 1 0^{- 2} - 2.58 \times 1 0^{- 2}

= 8.7 \times 1 0^{- 2} J

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

$8.7 \times 1 0^{- 2} J$

$12.2 \times 1 0^{- 2} J$

$8.7 \times 1 0^{- 1} J$

$12.2 \times 10^{- 1} J$

Q. When a spring is stretched by a distance x, it exerts a force, given by F=(−5x−16x3)N The work done, when the spring is stretched from 0.1,m to 0.2m is

8.7×10−2J

12.2×10−2J

8.7×10−1J

12.2×10−1J

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

$8.7 \times 1 0^{- 2} J$

$12.2 \times 1 0^{- 2} J$

$8.7 \times 1 0^{- 1} J$

$12.2 \times 10^{- 1} J$