Question

A block of mass 6kg hangs in equilibrium from a spring of spring constant 30 / . N m Find the energy stored in the spring

Answer 1

Given:

A block of mass 6kg hangs in equilibrium from a spring of spring constant 30 N/m.

To find:

Energy stored in the spring.

Diagram:

$\boxed{\setlength{\unitlength}{1cm}\begin{picture}(6,6)\put(2, 2){\line(0,1){2}}\put(2, 4){\line(1,0){2}}\put(4, 4){\line(0,-1){2}}\put(4, 2){\line(-1,0){2}}\put(3,3){\vector(0,-1){2}}\put(3,3){\vector(0,1){2}}\put(3,5.25){kx}\put(3,0.75){mg}\put(2,0){ \rm{\underline{FBD\:of\:Block}} }\end{picture}}$

Calculation:

Since the mass is in translational equilibrium, the spring force is equal and opposite to its weight.

Let the extension of spring be x ;

$\therefore \: kx = mg$

$= > 30x = 6g$

$= > 30x = 60$

$= > x = 2 \: m$

So, amount of spring energy

$\therefore \: E = \dfrac{1}{2} k {x}^{2}$

$= > \: E = \dfrac{1}{2} \times 30 \times {(2)}^{2}$

$= > \: E = \dfrac{1}{2} \times 30 \times 4$

$= > \: E = 30 \times 2$

$= > \: E = 60 \: joule$

So, final answer is:

$\boxed{ \sf{\: E = 60 \: joule}}$

Answer 2

Answer:

A block of mass 6kg hangs in equilibrium from a spring of spring constant 30 N/m, the energy stored in the spring is $54.1Nm$

Explanation:

• The elastic potential energy stored in the spring of spring constant (K) when displaced to a distance (x) is    

         $U=\frac{1}{2}kx^{2}$

• The restoring force on a spring of spring constant (K) when elongated to a distance (x) is given as $F=-Kx$

• Due to the mass (m) attached to the spring the gravitational force on the mass is  $F=mg$

        where $g=9.8m/s^{2}$(acceleration due to gravity)

• at equilibrium condition  both the forces are equal, that is $-Kx=mg$

From the question, we have

mass of the block attached to the spring $m=6kg$

the spring constant of the spring $K=30N/m$

stretched length $x=mg/K$

put the given values

⇒$x=\frac{6*9.8}{30} =1.96m$

now the energy stored is $U=\frac{30*1.9^{2} }{2} =54.1Nm$