An instrument to measure the acceleration of gravity on Mars is constructed of a spring from which is suspended a mass of 0.40 kg. At a place on earth where the local acceleration of gravity is 9.81 ms?, the spring extends 1.08 cm. When the instrument package is landed on Mars, it radios the information that the spring is extended 0.40 cm. What is the Martian acceleration of gravity?.
The correct answer and explanation is :
To determine the acceleration due to gravity on Mars, we will use Hooke’s Law and the relationship between force, mass, and gravity.
Step 1: Determine the Spring Constant (k)
Hooke’s Law states that the force exerted by a spring is given by: F=kxF = kx
where:
- FF is the force applied,
- kk is the spring constant,
- xx is the extension of the spring.
On Earth, the force acting on the mass is: F=mgF = mg
Given:
- m=0.40m = 0.40 kg,
- gEarth=9.81g_{\text{Earth}} = 9.81 m/s²,
- xEarth=1.08x_{\text{Earth}} = 1.08 cm = 0.0108 m,
k=Fx=(0.40)(9.81)0.0108=3.9240.0108=363.33 N/mk = \frac{F}{x} = \frac{(0.40)(9.81)}{0.0108} = \frac{3.924}{0.0108} = 363.33 \text{ N/m}
Step 2: Determine Gravity on Mars
On Mars, the same spring extends by xMars=0.40x_{\text{Mars}} = 0.40 cm = 0.0040 m. The force acting on the mass is: F=mgMarsF = mg_{\text{Mars}}
Since the same spring is used: mgMars=kxMarsmg_{\text{Mars}} = k x_{\text{Mars}}
Solving for gMarsg_{\text{Mars}}: gMars=kxMarsm=(363.33)(0.0040)0.40g_{\text{Mars}} = \frac{k x_{\text{Mars}}}{m} = \frac{(363.33)(0.0040)}{0.40} gMars=1.45330.40=3.63 m/s²g_{\text{Mars}} = \frac{1.4533}{0.40} = 3.63 \text{ m/s²}
Explanation
The spring stretches due to the force of gravity acting on the mass. Since the spring follows Hooke’s Law, the extension is proportional to the gravitational force. By comparing the extensions on Earth and Mars, and knowing the gravity on Earth, we can determine the unknown gravity on Mars. The acceleration due to gravity on Mars is found to be 3.63 m/s², which aligns closely with the accepted value (~3.7 m/s²).