Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 13

Sam pushes a \(10.0-\mathrm{kg}\) sack of bread flour on a frictionless horizontal surface with a constant horizontal force of \(2.0 \mathrm{N}\) starting from rest. (a) What is the kinetic energy of the sack after Sam has pushed it a distance of \(35 \mathrm{cm} ?\) (b) What is the speed of the sack after Sam has pushed it a distance of \(35 \mathrm{cm} ?\)

Short Answer

Expert verified
Question: A sack of bread flour with a mass of 10 kg is on a frictionless surface. A constant force of 2.0 N is used to push the sack a distance of 35 cm. Calculate (a) the kinetic energy of the sack after being pushed, and (b) the speed of the sack after being pushed. Answer: (a) The kinetic energy of the sack after being pushed for a distance of 35 cm is 0.7 J. (b) The speed of the sack after being pushed a distance of 35 cm is 0.37 m/s.

Step by step solution

01

Part (a): Kinetic Energy Calculation

We are given the initial kinetic energy \(K_{initial} = 0 \mathrm{J}\), the force applied \(F = 2.0 \mathrm{N}\) and the distance traveled \(d = 35 \mathrm{cm} = 0.35 \mathrm{m}\). We will use the work-energy principle, which states that the work done on an object is equal to its change in kinetic energy. The work done by the constant force can be calculated as follows: $$ W = F \cdot d $$ Once we find the work done, we can use it to find the final kinetic energy \(K_{final}\). Since the initial kinetic energy is zero, the final kinetic energy will be equal to the work done. $$ K_{final} = K_{initial} + W $$
02

Part (a): Calculate Work Done

Now we can calculate the work done using the given force and distance: $$ W = F \cdot d = (2.0 \mathrm{N}) \cdot (0.35 \mathrm{m}) = 0.7\ \mathrm{J} $$
03

Part (a): Calculate Final Kinetic Energy

Since the initial kinetic energy is zero, the final kinetic energy will be equal to the work done: $$ K_{final} = K_{initial} + W = 0 + 0.7\ \mathrm{J} = 0.7\ \mathrm{J} $$ The kinetic energy of the sack after being pushed for a distance of \(35 \mathrm{cm}\) is \(0.7\ \mathrm{J}\).
04

Part (b): Speed Calculation

Now we need to find the speed of the sack after being pushed for \(35\ \mathrm{cm}\). The kinetic energy (\(K\)) of an object is related to its mass (\(m\)) and its speed (\(v\)) by the following equation: $$ K = \frac{1}{2}mv^2 $$ Since we have found the final kinetic energy, we can use this equation to find the speed of the sack: $$ v = \sqrt{\frac{2K}{m}} $$
05

Part (b): Calculate Speed of the Sack

We will now calculate the speed of the sack using the final kinetic energy and its mass: $$ v = \sqrt{\frac{2 \cdot 0.7\ \mathrm{J}}{10.0\ \mathrm{kg}}} = \sqrt{\frac{1.4\ \mathrm{J}}{10.0\ \mathrm{kg}}} = \sqrt{0.14\ \mathrm{m^2/s^2}} = 0.37\ \mathrm{m/s} $$ The speed of the sack after being pushed a distance of \(35\ \mathrm{cm}\) is \(0.37\ \mathrm{m/s}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An automobile with a mass of \(1600 \mathrm{kg}\) has a speed of $30.0 \mathrm{m} / \mathrm{s} .$ What is its kinetic energy?
A skier starts from rest at the top of a frictionless slope of ice in the shape of a hemispherical dome with radius \(R\) and slides down the slope. At a certain height \(h,\) the normal force becomes zero and the skier leaves the surface of the ice. What is \(h\) in terms of \(R ?\)
A shooting star is a meteoroid that burns up when it reaches Earth's atmosphere. Many of these meteoroids are quite small. Calculate the kinetic energy of a meteoroid of mass \(5.0 \mathrm{g}\) moving at a speed of $48 \mathrm{km} / \mathrm{s}$ and compare it to the kinetic energy of a 1100 -kg car moving at \(29 \mathrm{m} / \mathrm{s}(65 \mathrm{mi} / \mathrm{h})\).
A 69.0 -kg short-track ice skater is racing at a speed of $11.0 \mathrm{m} / \mathrm{s}$ when he falls down and slides across the ice into a padded wall that brings him to rest. Assuming that he doesn't lose any speed during the fall or while sliding across the ice, how much work is done by the wall while stopping the ice skater?
The number of kilocalories per day required by a person resting under standard conditions is called the basal metabolic rate (BMR). (a) To generate 1 kcal, Jermaine's body needs approximately 0.010 mol of oxygen. If Jermaine's net intake of oxygen through breathing is 0.015 mol/min while he is resting, what is his BMR in kcal/day? (b) If Jermaine fasts for 24 h, how many pounds of fat does he lose? Assume that only fat is consumed. Each gram of fat consumed generates \(9.3 \mathrm{kcal}\)
See all solutions

Recommended explanations on Physics Textbooks

Circular Motion and Gravitation

Read Explanation

Geometrical and Physical Optics

Read Explanation

Electromagnetism

Read Explanation

Space Physics

Read Explanation

Modern Physics

Read Explanation

Rotational Dynamics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free