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Chapter 3: Problem 44

Then Jumps A \(29.0 \mathrm{~kg}\) child, with a \(4.50 \mathrm{~kg}\) backpack on his back, first stands on a sidewalk and then jumps up into the air. Find the magnitude and direction of the force on the sidewalk from the child when the child is (a) standing still and (b) in the air. Now find the magnitude and direction of the net force on Earth due to the child when the child is (c) standing still and (d) in the air.

Short Answer

Expert verified
Weight when standing: 330.75 N, downwards. Force on sidewalk when jumping: 0 N. Net force on Earth when standing: 330.75 N, upwards. Net force on Earth when in the air: 0 N.

Step by step solution

01

Calculate the total weight of the child and backpack (when standing still)

The total weight can be calculated using the formula: \[ W = (m_{child} + m_{backpack}) \times g \]Where \( m_{child} = 29.0 \mathrm{~kg} \) and \( m_{backpack} = 4.50 \mathrm{~kg} \), and \( g = 9.81 \mathrm{~m/s^2} \). Substitute these values into the formula to find the total weight.
02

Calculate the normal force on the sidewalk (when standing still)

Since the child is standing still, the force on the sidewalk is equal to the total weight calculated in Step 1.
03

Determine the direction of the force on the sidewalk (when standing still)

The force on the sidewalk due to the child standing still is directed downwards (towards the Earth). The normal force exerted by the sidewalk on the child will be equal in magnitude but upwards.
04

Velocity of the child (when jumping into the air)

When the child is jumping into the air, the contact with the sidewalk is momentarily zero, thus the force on the sidewalk from the child becomes zero at that instant.
05

Net force on Earth due to the child (when standing still)

Since the Earth and the child are in a system where forces are equal and opposite, the net force on Earth due to the child (when standing still) will be equal in magnitude to the total weight computed in Step 1 but opposite in direction (upwards).
06

Net force on Earth due to the child (when in the air)

When the child is in the air, there is no direct contact with the ground. Hence, the net force on Earth due to the child becomes zero at that moment.

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Key Concepts

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

force calculation
To solve any physics problem involving forces, we need to understand force calculation. A force results in the interaction between objects, that can cause an object with mass to change its velocity.
For this problem, we use the formula for weight (force due to gravity) given by:
For the child and backpack's total weight, we calculate:
Step 1:
Step 1:
Step Force calculation is key to solve similar physics questions.
Newton's third law
Newton's third law of motion states: 'For every action, there is an equal and opposite reaction.'
This means if object A exerts a force on object B, object B simultaneously exerts a force of equal magnitude in the opposite direction on object A.
In this problem, when the child stands on the sidewalk, the child exerts a downward force due to their weight, and the sidewalk exerts an equal but upward normal force.
Similarly, when the child jumps into the air, the force from the child on the sidewalk becomes zero, leading to no normal reaction force from the sidewalk at that instant.
This concept helps explain the interaction forces between the child and the sidewalk.
weight and mass
Understanding the distinction between weight and mass is crucial:
  • Mass is a measure of the amount of matter an object contains; it is measured in kilograms (kg).
  • Weight is a force resulting from the gravitational pull on an object's mass; it is calculated as:
The terms are often used interchangeably in everyday language, but they have distinct meanings in physics.
For example, your mass remains the same regardless of location, but your weight can change depending on the gravitational force acting on you.
In this problem, we calculate the child's total weight including the backpack by multiplying the combined mass with gravity.
normal force
The normal force is the support force exerted by a surface perpendicular to the object resting on it.
When the child is standing still on the sidewalk, the normal force equals the child鈥檚 weight, acting in the upward direction.
This normal force prevents the child from accelerating through the sidewalk.
In solving the problem, calculating this force allows us to understand the balance of forces when the child is at rest.
gravitational force
Gravitational force is the force of attraction between two masses.
For any object close to Earth鈥檚 surface, this force can be calculated as the product of the object鈥檚 mass and the acceleration due to gravity (In the given problem, the gravitational force on the child and backpack combined gives us their 'weight'.
Understanding gravitational force helps us solve for the normal force and the net force exerted on Earth by the child.
When the child is in the air, though the contact forces with the sidewalk vanish, the gravitational force still acts on the child.

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Most popular questions from this chapter

Your roommate peeks over your shoulder while you are reading a physics text and notices the following sentence: "In free fall the acceleration is always \(g\) and always straight downward regardless of the motion." Your roommate finds this peculiar and raises three objections: (a) If I drop a balloon or a feather, it doesn't fall nearly as fast as a brick. (b) Not everything falls straight down; if I throw a ball it can go sideways. (c) If I hold a wooden ball in one hand and a steel ball in the other, I can tell that the steel ball is being pulled down much more strongly than the wooden one. It will probably fall faster. How would you respond to these statements? Discuss the extent to which they invalidate the quoted statement. If they don't invalidate the statement, explain why.

When a nucleus captures a stray neutron, it must bring the neutron to a stop within the diameter of the nucleus by means of the strong force. That force, which "glues" the nucleus together, is approximately zero outside the nucleus. Suppose that a stray neutron with an initial speed of \(1.4 \times 10^{7} \mathrm{~m} / \mathrm{s}\) is just barely captured by a nucleus with diameter \(d=1.0 \times 10^{-14} \mathrm{~m}\). Assuming that the strong force on the neutron is constant, find the magnitude of that force. The neutron's mass is \(1.67 \times 10^{-27} \mathrm{~kg}\).

Suppose you have the following equipment available: an electronic balance, a motion detector and an electronic force sensor attached to a computer-based laboratory system. You would like to determine the mass of a block of ice that can slide smoothly along a very level table top without noticeable friction. (a) Describe how you would use some of the equipment to find the gravitational mass of the ice. (b) Describe how you would use some of the equipment to find the inertial mass of the ice. (c) Which of the two types of masses can be measured in outer space where gravitational forces are very small?

A parachutist bails out and freely falls \(50 \mathrm{~m}\). Then the parachute opens, and thereafter she slows at \(2.0 \mathrm{~m} / \mathrm{s}^{2}\). She reaches the ground with a speed of \(3.0 \mathrm{~m} / \mathrm{s}\). (a) How long is the parachutist in the air? (b) At what height does the fall begin?

The tension at which a fishing line snaps is commonly called the line's "strength." What minimum strength is needed for a line that is to stop a salmon of weight \(85 \mathrm{~N}\) in \(11 \mathrm{~cm}\) if the fish is initially drifting at \(2.8 \mathrm{~m} / \mathrm{s}\) ? Assume a constant acceleration.
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