9781118230725.pdf

5-3APPLYING NEWTON’S LAWS 113

Dead-End Solution: Let us now include force by writ-

ing, again for the xaxis,

FappFABmAa.

(We use the minus sign to include the direction of .)

BecauseFABis a second unknown, we cannot solve this

equation for a.

Successful Solution:Because of the direction in which force

is applied, the two blocks form a rigidly connected system.

We can relate the net force on the systemto the acceleration of

the systemwith Newton’s second law. Here, once again for the

xaxis, we can write that law as

Fapp(mAmB)a,

where now we properly apply to the system with

total mass mA mB. Solving for aand substituting known

values, we find

(Answer)

Thus, the acceleration of the system and of each block is in the

positive direction of the xaxis and has the magnitude 2.0 m/s^2.

(b) What is the (horizontal) force on block Bfrom

blockA(Fig. 5-18c)?

KEY IDEA

We can relate the net force on block Bto the block’s accel-

eration with Newton’s second law.

Calculation:Here we can write that law, still for compo-

nents along the xaxis, as

FBAmBa,

which, with known values, gives

FBA(6.0 kg)(2.0 m/s^2 )12 N. (Answer)

Thus, force is in the positive direction of the xaxis and

has a magnitude of 12 N.

F

:

BA

F

:

BA

a

Fapp

mAmB



20 N

4.0 kg6.0 kg

2.0 m/s^2.



F

:

app

F

:

app

F

:

AB

F

:

AB

Sample Problem 5.07 Acceleration of block pushing on block

Some homework problems involve objects that move to-

gether, because they are either shoved together or tied to-

gether. Here is an example in which you apply Newton’s

second law to the composite of two blocks and then to the

individual blocks.

In Fig. 5-18a, a constant horizontal force F of magni-

:

app

tude 20 N is applied to block Aof mass mA 4.0 kg, which

pushes against block Bof mass mB6.0 kg. The blocks slide

over a frictionless surface, along an xaxis.

(a) What is the acceleration of the blocks?

Serious Error: Because force is applied directly

to block A, we use Newton’s second law to relate that

force to the acceleration of block A. Because the motion

is along the xaxis, we use that law for xcomponents

(Fnet,xmax), writing it as

FappmAa.

However, this is seriously wrong because is not the

only horizontal force acting on block A. There is also the

force from block F B(Fig. 5-18b).

:

AB

F

:

app

:a

F

:

app



Figure 5-18 (a) A constant horizontal force is applied to block
A,which pushes against block B.(b) Two horizontal forces act on
blockA.(c) Only one horizontal force acts on block B.

F

:

app

FBA

(c)

x

B

(a)

x

A

B

Fapp

(b)

A x

Fapp FAB

This force causes the

acceleration of the full

two-block system.

This is the only force

causing the acceleration

of block B.

These are the two forces

acting on just block A.

Their net force causes

its acceleration.

Additional examples, video, and practice available at WileyPLUS

the upward acceleration is the 939 N reading on the scale. Thus,
the net force on the passenger is

FnetFNFg939 N708 N231 N, (Answer)

during the upward acceleration. However, his acceleration

ap,cabrelative to the frame of the cab is zero. Thus, in the non-

inertial frame of the accelerating cab,Fnetis not equal to

map,cab, and Newton’s second law does not hold.