An object experiencing a constant force accelerates at . What will the acceleration of this object be if

a. The force is doubled? Explain.

b. The mass is doubled?

c. The force is doubled and the mass is doubled?

Short Answer

Expert verified

(a) As the force is doubled, acceleration is doubled

(b) When the mass is doubled, the acceleration will be halved

(c) When both the force and the mass are doubled, the acceleration becomes,

a'=2F/2m=F/ma'=a

Step by step solution

01

Part (a) Step 1: Given Information

The acceleration of the object is 10m/s2.

02

Part (a) Step 2: Calculation

When the force is doubled, the acceleration will also double. The mass of the object is constant. Hence the acceleration value will increase to20m/s2.

03

Part (b) Step 1: Given information

The acceleration of the object is 10m/s2.

04

Part (b) Step 2: Calculation

Since mass and acceleration are inversely proportional to each other, hence when mass is increased, the acceleration will decrease. Therefore, as the mass is doubled, for the same force the acceleration will be reduced by half. Hence the new value of acceleration will be

5m/s2.
05

Part (c) Step 1: Given information

The acceleration of the object is 10m/s2.

06

Part (c) Step 2: Calculation

The force acting on an object is equal to the product of its mass and acceleration as F=ma, whereF is the force, m is the mass and a is the acceleration. Therefore, the expression for acceleration is a=F/m. When both the force and the mass are doubled, the acceleration becomes, a'=2F/2m=F/ma'=a

Hence under the given conditions, the acceleration remains unchanged and its value is again 10m/s2.

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.

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

You'll learn in Chapter 25 that the potential energy of two electric charges is inversely proportional to the distance between them. Two charges 30nmapart have 1.0Jof potential energy. What is their potential energy if they are 20nmapart?

Problems 42 through 52 describe a situation. For each, draw a motion diagram, a force-identification diagram, and a free-body diagram.

You are bungee jumping from a high bridge. You are moving

downward while the bungee cord is stretching.

A constant force is applied to an object, causing the object to accelerate at 10 m/s2. What will the acceleration be if

a. The force is halved?

b. The object’s mass is halved?

c. The force and the object’s mass are both halved?

d. The force is halved and the object’s mass is doubled?

Problems 35 through 40 show a free-body diagram. For each:

a. Identify the direction of the acceleration vector aand show it as a vector next to your diagram. Or, if appropriate, write a=0b. If possible, identify the direction of the velocity vector and show it as a labeled vector.

c. Write a short description of a real object for which this is the

correct free-body diagram. Use Examples 5.4, 5.5, and 5.6 as

models of what a description should be like.

Problems 35 through 40 show a free-body diagram. For each:

a. Identify the direction of the acceleration vector and show it as a vector next to your diagram. Or, if appropriate, write b. If possible, identify the direction of the velocity vector and show it as a labeled vector.

c. Write a short description of a real object for which this is the

correct free-body diagram. Use Examples 5.4, 5.5, and 5.6 as

models of what a description should be like.

See all solutions

Recommended explanations on Physics Textbooks

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