Question

A spring of spring constant 5 × 10³ N/m is stretched initially
by 5 cm from the unstretched position. Then the work
required to stretch it further by another 5 cm is
(a) 18.75 J (b) 25.00 J
(c) 6.25 J (d) 12.50 J

Answer 1

Answer:

b..

Explanation:

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Answer 2

Answer:

A spring of spring constant 5 × 10³ N/m is stretched initially  by 5 cm from the upstretched position. Then the work  required to stretch it further by

5 cm is $18.8J$

Explanation:

• The elastic potential energy is the energy stored when a force applied to deform an elastic object

• A spring with elastic constant K compressed or expanded at a distance x from its mean position, then the potential energy stored is

        $PE=\frac{1}{2}Kx^{2}$

• The work done by the spring is equal to the change in stored energy

        $W=\frac{1}{2}Kx^{2}$

From the question,

the spring constant $K=5*10^{3} N/m$

work done by the spring when it is stretched to $5cm$from normal length

$W_{1}=\frac{1}{2}5*10^{3} (0.05)^{2} =6.2J$

work done by the spring when it is stretched to $10cm$ from natural length

$W_{2}=\frac{1}{2}5*10^{3} (0.1)^{2} =25J$

work done by the spring when it stretched from $5cm$ to $10cm$

$W_{2} -W_{1}=(25-6.2)J=18.8J$