A spring constant is 200 N/m. If stretched by 0.1 m, find restoring force.

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Detailed Solution

Spring force for $k=200\ \text{N/m}$ $k$ $=$ $200$ $N/m$ , stretch $x=0.1\ \text{m}$ $x$ $=$ $0.1$ $m$
Answer: $F=kx=200\times0.1=20\ \text{N}$ $F$ $=$ $kx$ $=$ $200$ $\times$ $0.1$ $=$ $20$ $N$ toward equilibrium.
Method: Hooke’s law applies for small deformations where the spring is linear. The force is restoring, so direction is opposite displacement. Units: $\text{N}=\text{kg m s}^{-2}$ $N$ $=$ $kg m s$ $-2$ . If asked for elastic potential energy, use $U=\tfrac12 kx^2=\tfrac12\times200\times0.1^2=1\ \text{J}$ $U$ $=$ $21$ $$ $kx$ $2$ $=$ $21$ $$ $\times$ $200$ $\times$ $0.1$ $2$ $=$ $1$ $J$ . Check proportionality: doubling extension doubles force, quadruples energy. The sign convention is often omitted in magnitude questions; remember the vector points toward the mean position.