Question:
A block of $10\,kg$ mass moving on a frictionless surface with speed $5\,ms^{-1}$ compresses a spring by $5cm$ and comes to rest. The force constant of the spring (in $Nm^{-1}$) is:
A block of $10\,kg$ mass moving on a frictionless surface with speed $5\,ms^{-1}$ compresses a spring by $5cm$ and comes to rest. The force constant of the spring (in $Nm^{-1}$) is:
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Kinetic energy fully converts into spring potential on frictionless surface.
Updated On: Apr 24, 2026
- $2 \times 10^5$
- $2.5 \times 10^5$
- $3 \times 10^5$
- $1.5 \times 10^5$
- $1 \times 10^5$
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The Correct Option is
Solution and Explanation
Concept:
• Energy conservation: \[ \frac{1}{2}mv^2 = \frac{1}{2}kx^2 \]
Step 1: Given values
\[ m=10,\; v=5,\; x=5cm = 0.05m \]
Step 2: Substitute
\[ \frac{1}{2}(10)(5^2) = \frac{1}{2}k(0.05)^2 \] \[ 125 = \frac{1}{2}k(0.0025) \]
Step 3: Solve
\[ k = \frac{125 \times 2}{0.0025} = 100000 \] \[ = 1 \times 10^5 \] Final Conclusion:
Option (E)
• Energy conservation: \[ \frac{1}{2}mv^2 = \frac{1}{2}kx^2 \]
Step 1: Given values
\[ m=10,\; v=5,\; x=5cm = 0.05m \]
Step 2: Substitute
\[ \frac{1}{2}(10)(5^2) = \frac{1}{2}k(0.05)^2 \] \[ 125 = \frac{1}{2}k(0.0025) \]
Step 3: Solve
\[ k = \frac{125 \times 2}{0.0025} = 100000 \] \[ = 1 \times 10^5 \] Final Conclusion:
Option (E)
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