Question

the exact value of the acceleration due to gravity (g) varies with altitude and the makeup of the ground. suppose a spring scale is stretched 18 cm by a 100 - gram mass at sea level. the acceleration due to gravity atop mount everest is 0.3% smaller than at sea level. how will this affect the displacement measured by the spring scale? (1 point) because the mass is constant at 100 grams, the displacement will not change. the displacement will decrease by the square of 0.3, or 0.09%. because the value of g decreased, the displacement increased by the same amount, 0.3%. the displacement will also be 0.3% smaller. check answer remaining attempts : 3

Explanation:

Step1: Recall Hooke's Law

$F = kx$ and $F=mg$, so $mg=kx$, where $m$ is mass, $g$ is acceleration - due - to - gravity, $k$ is the spring constant, and $x$ is displacement.

Step2: Analyze the relationship

At sea - level, $m_1g_1 = kx_1$. At Mount Everest, $m_2g_2=kx_2$. Since $m_1 = m_2=m$ (mass is constant), we have $\frac{x_2}{x_1}=\frac{g_2}{g_1}$. Given $g_2=(1 - 0.003)g_1$.

Step3: Calculate the new displacement ratio

Substitute $g_2=(1 - 0.003)g_1$ into $\frac{x_2}{x_1}=\frac{g_2}{g_1}$, we get $\frac{x_2}{x_1}=0.997$. So $x_2$ is $0.3\%$ smaller than $x_1$.

Answer:

The displacement will also be 0.3% smaller.