An object attached to a spring is pulled across a horizontal frictionless surface. If the force constant (spring constant) of the spring is 45 N/m and the spring is stretched by 0.88 m when the object is accelerating at $1.4 \mathrm {~m} / \mathrm { s } ^ { 2 }$
what is the mass of the object?
A) 31 kg
B) 36 kg
C) 24 kg
D) 28 kg

Question

An object attached to a spring is pulled across a horizontal frictionless surface. If the force constant (spring constant) of the spring is  45 N/m and the spring is stretched by  0.88 m when the object is accelerating at $1.4 \mathrm {~m} / \mathrm { s } ^ { 2 }$
what is the mass of the object?
A)  31 kg
B)  36 kg
C)  24 kg
D)  28 kg

Quizplus · Accepted Answer

The force exerted by the spring can be calculated using Hooke's Law, $F = kx$, where $k = 45 \, \text{N/m}$ is the spring constant and $x = 0.88 \, \text{m}$ is the stretch. So, $F = 45 \times 0.88 = 39.6 \, \text{N}$. Using Newton's second law, $F = ma$, where $a = 1.4 \, \text{m/s}^2$ is the acceleration, we can find the mass $m$ by rearranging the equation to $m = \frac{F}{a} = \frac{39.6}{1.4} = 28.29 \, \text{kg}$, which rounds to $28 \, \text{kg}$.

An Object Attached to a Spring Is Pulled Across a Horizontal