A Block with mass of 0.1kg is attached to a spring and placed on a horizontal frictionless table the spring is stretched 20cm when a force of 5 newton is applied the spring constant ?
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F=kx
K=5/0.2=50/2=25N/M
T=2π✓m/K
T=2×3.142✓0.1/25
T=6.284×0.0632
T=0.397Sec
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6
Explanation:
25 newton / meter is the answer
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