If the potential energy of a spring is V on s | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(d) $U = \frac{1}{2}\left( {\frac{{YA}}{L}} \right)\,{l^2}$ $\Rightarrow$ $U \propto {l^2}$

$\frac{{{U_2}}}{{{U_1}}} = {\left( {\frac{{{l

Vedclass · Accepted Answer

$25V$

Vedclass · Answer

(d) $U = \frac{1}{2}\left( {\frac{{YA}}{L}} \right)\,{l^2}$ $\Rightarrow$ $U \propto {l^2}$

$\frac{{{U_2}}}{{{U_1}}} = {\left( {\frac{{{l_2}}}{{{l_1}}}} \right)^2} = {\left( {\frac{{10}}{2}} \right)^2} = 25$ $\Rightarrow$  ${U_2} = 25{U_1}$

i.e. potential energy of the spring will be $25 \,V$

If the potential energy of a spring is $V$ on stretching it by $2\, cm$, then its potential energy when it is stretched by $10 \,cm$ will be

Similar Questions

A wire suspended vertically from one of its ends is stretched by attaching a weight of $200\ N$ to the lower end. The weight stretches the wire by $1\ mm$. Then the elastic energy stored in the wire ......... $J$

If $x$ longitudinal strain is produced in a wire of Young's modulus $y,$ then energy stored in the material of the wire per unit volume is

On stretching a wire, the elastic energy stored per unit volume is

Identical springs of steel and copper are equally stretched. On which more work will have to be done ?

If one end of a wire is fixed with a rigid support and the other end is stretched by a force of $10 \,N,$ then the increase in length is $0.5\, mm$. The ratio of the energy of the wire and the work done in displacing it through $1.5\, mm$ by the weight is