A spring attached to the ceiling is stretched 2.45 meters by a four kilogram mass. The value of the Hooke's Law spring constant, $k$ , is
A) 1/4 meter-Newton
B) 4 meter-Newtons
C) 1/4 Newton per meter
D) 16 Newtons per meter
E) none of the above

Question

Quizplus · Accepted Answer

Hooke's Law is given by $F = kx$, where $F$ is the force applied, $k$ is the spring constant, and $x$ is the displacement of the spring from its equilibrium position. The force applied by the mass is due to gravity, $F = mg$, where $m = 4$ kg and $g \approx 9.8$ m/s$^2$. Therefore, $F = 4 \times 9.8 = 39.2$ N. Given $x = 2.45$ m, we solve for $k = F/x = 39.2 / 2.45 = 16$ Newtons per meter.

A Spring Attached to the Ceiling Is Stretched 2 kkk , Is A) 1/4 Meter-Newton B) 4 Meter-Newtons C) 1/4

A Spring Attached to the Ceiling Is Stretched 2 $k$ , Is
A) 1/4 Meter-Newton
B) 4 Meter-Newtons
C) 1/4