Question:
A person of mass 80 kg sits on a spring and compresses it by 2 m. If the person is projected at 20 m/s velocity by the spring, then the spring constant is (Take no energy loss):
A person of mass 80 kg sits on a spring and compresses it by 2 m. If the person is projected at 20 m/s velocity by the spring, then the spring constant is (Take no energy loss):
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Always use energy conservation when spring compression is directly related to velocity.
Updated On: Jun 19, 2026
- 2000
- 4000
- 6000
- 8000
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The Correct Option is D
Solution and Explanation
Step 1: Understand energy conversion.
The energy stored in the compressed spring is converted into kinetic energy of the person. Thus: \[ \frac{1}{2}kx^2 = \frac{1}{2}mv^2 \]
Step 2: Given values.
\[ m = 80\,kg,\quad x = 2\,m,\quad v = 20\,m/s \]
Step 3: Substitute values in equation.
\[ \frac{1}{2}k(2)^2 = \frac{1}{2}(80)(20)^2 \]
Step 4: Simplify both sides.
Left side: \[ 2k \] Right side: \[ \frac{1}{2} \cdot 80 \cdot 400 = 16000 \]
Step 5: Solve for k.
\[ 2k = 16000 \Rightarrow k = 8000 \, N/m \]
Step 6: Final conclusion.
Thus, the spring constant is: \[ \boxed{8000 \, N/m} \]
The energy stored in the compressed spring is converted into kinetic energy of the person. Thus: \[ \frac{1}{2}kx^2 = \frac{1}{2}mv^2 \]
Step 2: Given values.
\[ m = 80\,kg,\quad x = 2\,m,\quad v = 20\,m/s \]
Step 3: Substitute values in equation.
\[ \frac{1}{2}k(2)^2 = \frac{1}{2}(80)(20)^2 \]
Step 4: Simplify both sides.
Left side: \[ 2k \] Right side: \[ \frac{1}{2} \cdot 80 \cdot 400 = 16000 \]
Step 5: Solve for k.
\[ 2k = 16000 \Rightarrow k = 8000 \, N/m \]
Step 6: Final conclusion.
Thus, the spring constant is: \[ \boxed{8000 \, N/m} \]
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