/*! 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} Q95P Suppose that in Fig. 5-12, the m... [FREE SOLUTION] | 91影视

91影视

Suppose that in Fig. 5-12, the masses of the blocks are 2.0 kg and 4.0 kg. (a) Which mass should the hanging block have if the magnitude of the acceleration is to be as large as possible? What then are (b) the magnitude of the acceleration and (c) the tension in the cord?

Fig. 5-12

Short Answer

Expert verified
  1. Hanging mass is 4kg
  2. Acceleration is 6.5m/s2
  3. Tension T = 13N

Step by step solution

01

Given

m1=2kgm2=4kg

02

Understanding the concept

Using the concept of Newton鈥檚 second law of motion, we can decide which mass should be hanging. We can also write two equations for two masses in terms of tension and acceleration, i.e. two unknown variables. Tension and acceleration are common for both the objects. So these two equations can be solved for the acceleration and tension in the string.

Formula

Fnet=ma

Here,Fnet is the force, m is the mass of the object, and a is the acceleration of the object.

03

Draw the Free Body Diagram

04

Find out which mass should the hanging block have if the magnitude of the acceleration is to be as large as possible 

(a)

Heavier mass should be hanging for maximum acceleration as net force is greater for heavier mass, so 4 mass should be hanging.

05

Calculate the magnitude of the acceleration

(b)

Net force on mass m1is as follow

Fnet=T

But according to newton鈥檚 second law,

Fnet=m1a

So

role="math" localid="1660911650209" T=m1a

Now net force on second massm2=2kg

m2g-T=m2am2g-m1a=m2a

Rearrange this equation to find acceleration.

a=m2gm1+m2

Substitute the values in the above equation to calculate acceleration

a=4kg9.811m/s22kg+4kg=6.54m/s2

06

Calculate the tension in the chord

(c)

Tension in chord is calculated using Newton鈥檚 law as,

T=m1a=26.54m/s2=13.08N

Therefore, the tension in the cord is 13 N.

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 91影视!

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

A car traveling at 53Km/hhits a bridge abutment. A passenger in the car moves forward a distance of65cm(with respect to the road) while being brought to rest by an inflated air bag. What magnitude of force (assumed constant) acts on the passenger鈥檚 upper torso, which has a mass of41kg?

In Fig. 5-64, a forceof magnitude 12Nis applied to a FedEx box of mass m2=1.0kg. The force is directed up a plane tilted by =37. The box is connected by a cord to a UPS box of mass m1=3.0kgon the floor. The floor, plane, and pulley are friction less, and the masses of the pulley and cord are negligible. What is the tension in the cord?

A 40KGskier skis directly down a frictionless slope angled at 100to the horizontal. Assume the skier moves in the negative direction of an xaxis along the slope. A wind force with component role="math" localid="1657169142011" Fxacts on the skier. (a) What is Fxif the magnitude of the skier鈥檚 velocity is constant, (b) What isif the magnitude of the skier鈥檚 velocity is increasing at a rate of 1.0m/s2, and (c) What is Fxif the magnitude of the skier鈥檚 velocity is increasing at a rate of2.0m/s2?

The velocity of a3.00kgparticle is given byv=(8.00+2)m/swith time tin seconds. At the instant the net force on the particle has a magnitude of35.0N, (a) what are the direction (relative to the positive direction of the xaxis) of the net force and (b) what are the direction (relative to the positive direction of the xaxis) of the particle鈥檚 direction of travel?

The tension at which a fishing line snaps is commonly called the line鈥檚 鈥渟trength.鈥 What minimum strength is needed for a line that is to stop a salmon of weight 85 Nin 11 cmif the fish is initially drifting at2.8m/s? Assume a constant deceleration.

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.