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A spring of spring constant 5 × 103 is stretched initially by 5 cm from the unstretched position. The work required to stretch it further by another 5 cm is
- a)6025 N-m
- b)12.50 N-m
- c)18.75 N-m
- d)25.00n-m
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A spring of spring constant 5× 103 is stretched initially by 5 c...
Spring constant, k = 5 × 103 N/m
The work required to stretch the spring by the displacement x is W = 1/2 kx2
When the spring is stretched by 5 cm, work done is W1 = 1/2 × (5 × 103N/m) × (5 × 10-2m)2
W1 = 6.25J
When the spring is again stretched further by 5 cm, the net displacement of the spring from its equilibrium position is
x' = 5 cm, +5 cm = 10 cm
Thus, the work required to stretch it further by an additional 5 cm is W2 = 1/2 x (5 × 103N/m) × (10 × 10-2m)2
⇒ W2 = 25J
So, the extra work required to stretch the spring by an additional
5 cmis ΔW = W2 - W1 ⇒ ΔW = 25J - 6.25J ⇒ ΔW = 18.75J or 18.75 N - m
The work required to stretch the spring by the displacement x is W = 1/2 kx2
When the spring is stretched by 5 cm, work done is W1 = 1/2 × (5 × 103N/m) × (5 × 10-2m)2
W1 = 6.25J
When the spring is again stretched further by 5 cm, the net displacement of the spring from its equilibrium position is
x' = 5 cm, +5 cm = 10 cm
Thus, the work required to stretch it further by an additional 5 cm is W2 = 1/2 x (5 × 103N/m) × (10 × 10-2m)2
⇒ W2 = 25J
So, the extra work required to stretch the spring by an additional
5 cmis ΔW = W2 - W1 ⇒ ΔW = 25J - 6.25J ⇒ ΔW = 18.75J or 18.75 N - m
Most Upvoted Answer
A spring of spring constant 5× 103 is stretched initially by 5 c...
Given:
Spring constant, k = 5 × 10^3 N/m
Initial stretch, x1 = 5 cm = 0.05 m
Calculating work done in stretching by 5 cm:
Work done in stretching a spring is given by the formula:
W = 0.5 * k * (x2^2 - x1^2)
where x2 is the final stretch.
Given x1 = 0.05 m and x2 = 0.1 m (5 cm further stretch)
Plugging in the values:
W = 0.5 * 5 × 10^3 * ((0.1)^2 - (0.05)^2)
W = 0.5 * 5 × 10^3 * (0.01 - 0.0025)
W = 0.5 * 5 × 10^3 * 0.0075
W = 18.75 N-m
Therefore, the work required to stretch the spring further by another 5 cm is 18.75 N-m, which corresponds to option 'c'.
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A spring of spring constant 5× 103 is stretched initially by 5 cmfrom the unstretched position. The work required to stretch it further by another 5 cm isa)6025 N-mb)12.50 N-mc)18.75 N-md)25.00n-mCorrect answer is option 'C'. Can you explain this answer?
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