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On a planet far, far away, an astronaut picks up a rock. The rock has a mass of \(5.00 \mathrm{kg}\), and on this particular planet its weight is \(40.0 \mathrm{N}\). If the astronaut exerts an upward force of \(46.2 \mathrm{N}\) on the rock, what is its acceleration?

Short Answer

Expert verified
The acceleration of the rock is \(1.24 \mathrm{m/s^2}\).

Step by step solution

01

Understand the problem

We need to find the acceleration of a rock when an upward force is applied. The rock's mass is given, the weight on the planet is given, and the force applied is given.
02

Calculate gravitational acceleration on the planet

Use the weight formula, where weight (W) is the product of mass (m) and gravitational acceleration (g): \( W = m \times g \). Given the weight \( W = 40.0 \mathrm{N} \) and mass \( m = 5.00 \mathrm{kg} \), solve for \( g \) as follows: \( g = \frac{W}{m} = \frac{40.0 \mathrm{N}}{5.00 \mathrm{kg}} = 8.0 \mathrm{m/s^2} \).
03

Net force calculation

The net force (\(F_{net}\)) on the rock can be determined by subtracting the weight of the rock from the upward force applied: \( F_{net} = F_{upward} - W \). Plugging the values, \( F_{net} = 46.2 \mathrm{N} - 40.0 \mathrm{N} = 6.2 \mathrm{N} \).
04

Calculate acceleration using Newton's Second Law

Newton's Second Law states that \( F_{net} = m \times a \). Here \( F_{net} = 6.2 \mathrm{N} \) and \( m = 5.00 \mathrm{kg} \). Solve for acceleration \( a \): \( a = \frac{F_{net}}{m} = \frac{6.2 \mathrm{N}}{5.00 \mathrm{kg}} = 1.24 \mathrm{m/s^2} \).

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Key Concepts

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

Gravitational Acceleration
Gravitational acceleration is a measure of how quickly an object speeds up when it is falling freely under gravity alone. This quantity varies depending on the celestial body you are on. Unlike Earth, other planets have different masses and radii, which influence their gravitational pull.
Gravitational acceleration can be derived from the weight of an object, as illustrated in this exercise. The weight of an object is the force exerted on it due to gravity, calculated by multiplying the mass of the object by the gravitational acceleration, using the formula:
  • \( W = m \times g \)
Here, \( g \) represents the gravitational acceleration.
For the given problem, we know the weight of the rock is 40.0 N and the mass is 5.00 kg. By arranging the equation to solve for \( g \), we have:
  • \( g = \frac{W}{m} = \frac{40.0 \; \mathrm{N}}{5.00 \; \mathrm{kg}} = 8.0 \; \mathrm{m/s^2} \)
This calculation informs us that the gravitational acceleration on the distant planet is 8.0 m/s², which is different from Earth's 9.81 m/s².
Net Force
The concept of net force is crucial in understanding how objects move under various influences. Net force is essentially the sum of all forces acting on an object and determines how the object's motion changes.
In our example, the astronaut applies an upward force of 46.2 N to counteract the rock's weight. The weight acts downwards with a force of 40.0 N. To find the net force, we subtract the downward force (weight) from the upward force exerted by the astronaut:
  • \( F_{net} = F_{upward} - W = 46.2 \; \mathrm{N} - 40.0 \; \mathrm{N} = 6.2 \; \mathrm{N} \)
The resulting net force of 6.2 N indicates that there is still an upward force acting on the rock after overcoming the planet's gravitational pull. This net force is responsible for the rock's acceleration.
Acceleration Calculation
Acceleration is the rate of change of velocity of an object when a net force is applied, as expressed in Newton's Second Law. This law is succinctly written as:
  • \( F_{net} = m \times a \)
where \( F_{net} \) is the net force, \( m \) is the mass, and \( a \) is the acceleration we are looking to find.
In the present exercise, given the net force of 6.2 N and the rock's mass of 5.00 kg, we rearrange the formula to solve for acceleration:
  • \( a = \frac{F_{net}}{m} = \frac{6.2 \; \mathrm{N}}{5.00 \; \mathrm{kg}} = 1.24 \; \mathrm{m/s^2} \)
This shows that the rock accelerates upwards at a rate of 1.24 m/s² when the astronaut applies a force of 46.2 N on the planet. It highlights how a net force results in a change in the rock's velocity.

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Most popular questions from this chapter

\- CE Predict/Explain A small car collides with a large truck. (a) Is the magnitude of the force experienced by the car greater than, less than, or equal to the magnitude of the force experienced by the truck? (b) Choose the best explanation from among the following: I. Action-reaction forces always have equal magnitude. II. The truck has more mass, and hence the force exerted on it is greater. III. The massive truck exerts a greater force on the lightweight car.

IP Responding to an alarm, a \(102-\mathrm{kg}\) fireman slides down a pole to the ground floor, 3.3 m below. The fireman starts at rest and lands with a speed of \(4.2 \mathrm{m} / \mathrm{s}\). (a) Find the average force exerted on the fireman by the pole. (b) If the landing speed is half that in part (a), is the average force exerted on the fireman by the pole doubled? Explain. (c) Find the average force exerted on the fireman by the pole when the landing speed is \(2.1 \mathrm{m} / \mathrm{s}\).

IP While waiting at the airport for your flight to leave, you observe some of the jets as they take off. With your watch you find that it takes about 35 seconds for a plane to go from rest to takeoff speed. In addition, you estimate that the distance required is about \(1.5 \mathrm{km}\). (a) If the mass of a jet is \(1.70 \times 10^{5} \mathrm{kg}\) what force is needed for takeoff? (b) Describe the strategy you used to solve part (a).

IP A gardener mows a lawn with an old-fashioned push mower. The handle of the mower makes an angle of \(35^{\circ}\) with the surface of the lawn. (a) If a \(219-\mathrm{N}\) force is applied along the handle of the \(19-\mathrm{kg}\) mower, what is the normal force exerted by the lawn on the mower? (b) If the angle between the surface of the lawn and the handle of the mower is increased, does the normal force exerted by the lawn increase, decrease, or stay the same? Explain.

An air-track cart of mass \(m_{1}=0.14 \mathrm{kg}\) is moving with a speed \(v_{0}=1.3 \mathrm{m} / \mathrm{s}\) to the right when it collides with a cart of mass \(\mathrm{m}_{2}=0.25 \mathrm{kg}\) that is at rest. Each cart has a wad of putty on its bumper, and hence they stick together as a result of their collision. Suppose the average contact force between the carts is \(F=1.5 \mathrm{N}\) during the collision. (a) What is the acceleration of cart \(1 ?\) Give direction and magnitude. (b) What is the acceleration of cart \(2 ?\) Give direction and magnitude. (c) How long does it take for both carts to have the same speed? (Once the carts have the same speed the collision is over and the contact force vanishes.) (d) What is the final speed of the carts, \(v_{\mathrm{f}} ?\) (e) Show that \(m_{1}{ }^{2} v_{0}\) is equal to \(\left(m_{1}+m_{2}\right) v_{1}\). (We shall investigate the significance of this result in Chapter 9.3

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