What is free-fall acceleration on the surface of the moon?

1 answer:

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R is the radius of the object. There is a negative sign in front of the equation because objects in free fall always fall downwards toward the center of the object. The acceleration due to gravity is 1.62 m/s2. This is approximately 1/6 that of the acceleration due to gravity on Earth, 9.81 m/s2.

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julia-pushkina [17]

Answer:

Approximately $8.4 \times 10^{2}\; \rm N$ , assuming that $g = 9.8\; \rm m \cdot s^{-2}$ .

Explanation:

Let $m$ and $a$ denote the mass and acceleration of Spiderman, respectively.

There are two forces on Spiderman:

Downward gravitational attraction from the earth: $W = m \cdot g$ .
Upward tension force from the strand of web $F(\text{tension})$ .

The directions of these two forces are exactly opposite of one another. Besides, because Spiderman is accelerating upwards, the magnitude of $F(\text{tension})$ (which points upwards) should be greater than that of $W$ (which points downwards towards the ground.)

Subtract the smaller force from the larger one to find the net force on Spiderman:

$(\text{Net Force}) = F(\text{tension}) - W$ .

On the other hand, apply Newton's Second Law of motion to find the value of the net force on Spiderman:

$(\text{Net Force}) = m \cdot a$ .

Combine these two equations to get:

$m \cdot a = (\text{Net Force}) = F(\text{tension}) - W$ .

Therefore:

$\begin{aligned}& F(\text{tension})\\ &= m \cdot a + W \\ &= m \cdot (a + g)\\ &= 76\; \rm kg \times \left(1.3\; \rm m \cdot s^{-2} + 9.8\; \rm m \cdot s^{-2}\right)\\ &\approx 8.4\times 10^{2}\; \rm N\end{aligned}$ .

By Newton's Third Law of motion, Spiderman would exert a force of the same size on the strand of web. Hence, the size of the force in the strand of the web should be approximately $8.4\times 10^{2}\; \rm N$ (downwards.)