/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q7-12Q Figure 7-23 shows three arrangem... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Figure 7-23 shows three arrangements of a block attached to identical springs that are in their relaxed state when the block is centered as shown. Rank the arrangements according to the magnitude of the net force on the block, largest first, when the block is displaced by distance d (a) to the right and (b) to the left. Rank the arrangements according to the work done on the block by the spring forces, greatest first, when the block is displaced by d (c) to the right and (d) to the left.

Short Answer

Expert verified
  1. Rank according to the net force acting on block it is displaced to the right is 3 > 2 > 1.
  2. Rank according to the net force acting on block it is displaced to the left is 3 > 2 > 1.
  3. Rank according to the work done on block when it is displaced to the right is 3 > 2 > 1.
  4. Rank according to the work done on block when it is displaced to the left is 3 > 2 > 1.

Step by step solution

01

The given data

Figure shows the three arrangements of a block attached to identical springs is shown.

02

Understanding the concept of the force and the work done

Using the concept of the net force acting on a body attached to the spring, we can calculate the value of the ranks of the blocks' positions according to their forces for three identical springs.

Formulae:

The force acting on the block due to attached spring,

F=-k.x (i)

The work done due to an applied force,

W=F.d (ii)

03

Calculation of the ranks of the situations due to force when block is displaced to the right

a)

When the spring is displaced by x in horizontal direction, then the spring force is given by equation (i).

For situation 1:

Here, two identical springs are attached in the system, so the net force is given using equation (i) as:

F1net=Fs1+Fs1=2Fs1=-2k.x

For situation 2:

Here, three identical springs are attached in the system, so the net force is given using equation (i) as follows:

F3net=Fs1+Fs1+Fs1=3Fs1=-3k.x

For situation 3:

Here, four identical springs are attached in the system, so the net force is given using equation (i) as:

F1net=Fs1+Fs1=Fs1=4Fs1=-4k.x

So, the rank for magnitude for force for both the directions is 3 > 2 > 1

Hence, the rank when block is displaced by the right is 3 > 2 > 1.

04

Calculation of the ranks of the situations due to force when block is displaced to the left

b)

From the calculations of part (a), we get that the rank is similar to both the directions.

Hence, the rank when block is displaced by the left is 3 . 2 > 1.

05

Calculation of the ranks of the situations due to work done when block is displaced to the right

c)

Work is done due to spring force. As displacement is the same for all cases, rank of the work is the same as the rank for spring force using equation (ii).

So, rank for the work done for both the directions is 3 > 2 > 1

Hence, the rank when block is displaced to the right is 3 > 2 > 1 .

06

Calculation of the ranks of the situations due to work done when block is displaced to the left

d)

From the above calculations of part (c), we get that the rank for both the directions is same.

Hence, the rank when block is displaced to the left is 3 > 2 > 1.

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 91Ó°ÊÓ!

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

Boxes are transported from one location to another in a warehouse by means of a conveyor belt that moves with a constant speed of 0.50 m/s. At a certain location the conveyor belt moves for 2.0 m up an incline that makes an angle of 10°with the horizontal, then for 2.0 m horizontally, and finally for 2.0 m down an incline that makes an angle of 10°ith the horizontal. Assume that a 2.0 kg box rides on the belt without slipping. At what rate is the force of the conveyor belt doing work on the box as the box moves (a) up the 10°incline, (b) horizontally, and (c) down the 10°incline?

Figure 7-27 shows an overhead view of three horizontal forces acting on a cargo canister that was initially stationary but now moves across a frictionless floor. The force magnitudes are F1=3.00N, F2=4.00N, and F3=10.00N, and the indicated angles are θ2=50.0°and θ3=35.0°. What is the net work done on the canister by the three forces during the first 4.00 m of displacement?

A coin slides over a frictionless plane and across an xy coordinate system from the origin to a point with xy coordinates (3.0m,4.0m) while a constant force acts on it. The force has magnitude 2.0 Nand is directed at a counterclockwise angle of100° from the positive direction of the x axis. How much work is done by the force on the coin during the displacement?

A bead with mass 1.8×10-2kg is moving along a wire in the positive direction of an x axis. Beginning at time t=0, when the bead passes through x=0 with speed 12 m/s, a constant force acts on the bead. Figure 7-24 indicates the bead’s position at these four times: t0=0, t1=1.0 s, t2=2.0 s, and t3=3.0 s. The bead momentarily stops at t=3.0 s. What is the kinetic energy of the bead at t=10 s ?

A 3.0 kg body is at rest on a frictionless horizontal air track when a constant horizontal force F→acting in the positive direction of an x axis along the track is applied to the body. A stroboscopic graph of the position of the body as it slides to the right is shown in Fig. 7-25. The force F→is applied to the body at t=0, and the graph records the position of the body at 0.50 s intervals. How much work is done on the body by the applied force F→between t=0 and t=2.0 s ?

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.