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What does a 138 \(\mathrm{N}\) rock weigh if it is accelerating (a) upward at \(12 \mathrm{m} / \mathrm{s}^{2},\) (b) downward at 3.5 \(\mathrm{m} / \mathrm{s}^{2} ?\) (c) What would be the answers to parts (a) and (b) if the rock had a mass of 138 \(\mathrm{kg} ?(\mathrm{d})\) What would be the answers to parts (a) and (b) if the rock were moving with a constant upward velocity of 23 \(\mathrm{m} / \mathrm{s} ?\)

Short Answer

Expert verified
(a) 306.94 N, (b) 88.70 N, (c) 3008.4 N and 869.4 N, (d) 138 N each.

Step by step solution

01

Understanding the Problem

We have to find the effective weight of the rock when it accelerates upwards and downwards. The effective weight might be different from the true weight depending on the acceleration applied.
02

Analyze Part (a)

For part (a), the rock is accelerating upward with an acceleration of 12 m/s^2. The formula for effective weight is given by:\[ F = m \cdot (g + a) \]where \( g = 9.8 \text{ m/s}^2 \) is the acceleration due to gravity and \( a = 12 \text{ m/s}^2 \) is the upward acceleration. Here, weight given is a force of 138 N, so use:\[ m = \frac{138}{g} \] to find mass, then substitute in the formula.
03

Calculate Mass

The mass of the rock can be calculated using its weight and gravitational force:\[ m = \frac{138}{9.8} \approx 14.08 \text{ kg}\]
04

Calculate Effective Weight for Upward Acceleration

Now we substitute the values into the formula:\[ F = 14.08 \cdot (9.8 + 12) = 14.08 \cdot 21.8 \approx 306.94 \text{ N} \]
05

Analyze Part (b)

For part (b), the rock is accelerating downward with an acceleration of 3.5 m/s^2. Use the formula for effective weight:\[ F = m \cdot (g - a) \]Substitute the known values to find the weight.
06

Calculate Effective Weight for Downward Acceleration

Substituting the given values we have:\[ F = 14.08 \cdot (9.8 - 3.5) = 14.08 \cdot 6.3 \approx 88.70 \text{ N} \]
07

Analyze Part (c) with Mass 138 kg

Change the condition to use a mass of 138 kg and repeat the calculations for both upward and downward accelerations.
08

Calculate for 138 kg Upward

For upward:\[ F = 138 \cdot (9.8 + 12) = 138 \cdot 21.8 = 3008.4 \text{ N} \]
09

Calculate for 138 kg Downward

For downward:\[ F = 138 \cdot (9.8 - 3.5) = 138 \cdot 6.3 = 869.4 \text{ N} \]
10

Analyze Part (d) Constant Velocity

If the rock moves with a constant velocity, its acceleration is zero. Therefore, effective forces for upward and downward motion would both be 138 N, as they rely solely on gravity in this scenario.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Weight Calculation
The weight of an object is the force with which it is pulled towards the Earth due to gravity. It is calculated using the formula:
  • \( W = m \cdot g \)
where \( m \) is the mass of the object and \( g = 9.8 \text{ m/s}^2 \) is the acceleration due to gravity.
It is important to remember that weight is a force and is measured in Newtons (N).
Thus, to calculate the weight of an object like a rock, you need to know its mass. For example, if a rock has a weight of 138 N, you can find its mass by rearranging the formula to \( m = \frac{W}{g} \).
This calculation shows that the mass of the rock is approximately 14.08 kg.
Effective Weight
Effective weight refers to the way your body feels, depending on whether you're accelerating or decelerating. It changes depending on movement and the forces acting on the object.
Effective weight can be more or less than the actual gravitational force due to additional accelerations.
  • When you accelerate upwards, like a rock being thrown upwards, your effective weight feels heavier due to additional forces.
  • Conversely, when there is downward acceleration, it feels lighter.
To calculate effective weight under different accelerations, use the equation:
  • \( F_{\text{effective}} = m \cdot (g + a) \) for upward acceleration, and
  • \( F_{\text{effective}} = m \cdot (g - a) \) for downward acceleration.
This ensures you account for the changes brought about by acceleration.
Acceleration Effects
Acceleration influences how heavy or light you feel, creating an effect on your perceived weight. When you're accelerating:
  • Upwards, the acceleration adds to the gravitational pull, making the object feel heavier.
  • Downwards, it subtracts from gravitational pull, causing a lighter feeling.
For example, if a rock usually under the force of gravity accelerating up at 12 m/s², the effective force of weight becomes greater. Simply calculate using:
  • \( F = m \cdot (g + a) \)
With a downward acceleration, this works differently, as you subtract the outside acceleration from gravitational force:
  • \( F = m \cdot (g - a) \)
Therefore, understanding these principles clarifies how movement affects weight sensation.
Gravitational Force
Gravitational force is the natural force exerted by the Earth pulling objects towards its center. It keeps objects anchored and creates the sensation of weight. Gravity provides a constant acceleration of 9.8 \( \text{m/s}^2 \).
You experience this force uniformly, whether moving or at rest.
It is important to recognize that gravity remains constant while acceleration changes. The gravitational force acts as the baseline upon which to calculate effective weight when additional forces like acceleration are present.
When the question involves movement or different forces, we adjust our calculations for these conditions but always start with the gravitational factor. Understanding gravity's role makes it easier to predict changes in effective weight under different circumstances.

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Most popular questions from this chapter

The space shuttle. During the first stage of its launch, a space shuttle goes from rest to 4973 \(\mathrm{km} / \mathrm{h}\) while rising a vertical distance of 45 \(\mathrm{km}\) . Assume constant acceleration and no variation in \(g\) over this distance. (a) What is the acceleration of the shuttle? (b) If a 55.0 \(\mathrm{kg}\) astronaut is standing on a scale inside the shuttle during this launch, how hard will the scale push on her? Start with a free-body diagram of the astronaut. (c) If this astronaut did not realize that the shuttle had left the launch pad, what would she think were her weight and mass?

BIO Bacterial motion. A bacterium using its flagellum as propulsion can move through liquids at a rate of 0.003 \(\mathrm{m} / \mathrm{s} .\) For a \(50 \mu \mathrm{m}-\) long bacterium, that is the equivalent of 60 cell lengths per second. Bacteria of that size have a mass of approximately \(1 \times 10^{-12} \mathrm{g}\). The viscous drag on a swimming bacterium is so great that if it stops beating its flagellum it will stop within a distance of 0.01 \(\mathrm{nm} .\) What is the acceleration that stops the bacterium? A. \(1.2 \times 10^{4} \mathrm{m} / \mathrm{s}^{2}\) B. \(5 \times 10^{5} \mathrm{m} / \mathrm{s}^{2}\) C. \(6 \times 10^{5} \mathrm{m} / \mathrm{s}^{2}\) D. \(9 \times 10^{5} \mathrm{m} / \mathrm{s}^{2}\)

A parachutist relies on air resistance (mainly on her parachute) to decrease her downward velocity. She and her parachute have a mass of 55.0 \(\mathrm{kg}\) , and at a particular moment air resistance exerts a total upward force of 620 \(\mathrm{N}\) on her and her parachute. (a) What is the weight of the parachutist? \((\mathrm{b})\) Draw a free-body diagram for the parachutist (see Section \(4.6 ) .\) Use that diagram to calculate the net force on the parachutist. Is the net force upward or downward? (c) What is the acceleration (magnitude and direction) of the parachutist?

An astronaut's pack weighs 17.5 \(\mathrm{N}\) when she is on earth but only 3.24 \(\mathrm{N}\) when she is at the surface of an asteroid. (a) What is the acceleration due to gravity on this asteroid? (b) What is the mass of the pack on the asteroid?

Planet \(\mathbf{X} !\) When venturing forth on Planet \(X,\) you throw a 5.24 kg rock upward at 13.0 \(\mathrm{m} / \mathrm{s}\) and find that it returns to the same level 1.51 \(\mathrm{s}\) later. What does the rock weigh on Planet \(\mathrm{X}\) ?

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