Calculate the force the professor exerts on the cart in Figure 2 using data from the previous example if needed.

*Strategy*

If we now define the system of interest to be the cart plus equipment (System 2 in Figure 2), then the net external force on System 2 is the force the professor exerts on the cart minus friction. The force she exerts on the cart, FprofFprof size 12{F rSub { size 8{"prof"} } } {}, is an external force acting on System 2. FprofFprof size 12{F rSub { size 8{"prof"} } } {} was internal to System 1, but it is external to System 2 and will enter Newton’s second law for System 2.

*Solution*

Newton’s second law can be used to find FprofFprof size 12{F rSub { size 8{"prof"} } } {}. Starting with

a
=
F
net
m
a
=
F
net
m
size 12{a = { {F rSub { size 8{"net"} } } over {m} } } {}

(5)and noting that the magnitude of the net external force on System 2 is

Fnet=Fprof−f,Fnet=Fprof−f size 12{F rSub { size 8{"net"} } = F rSub { size 8{"prof"} } -f} {},

(6)we solve for FprofFprof size 12{F rSub { size 8{"prof"} } } {}, the desired quantity:

F
prof
=
F
net
+
f
.
F
prof
=
F
net
+
f
.
size 12{F rSub { size 8{"prof"} } = F rSub { size 8{"net"} } + f} {}

(7)The value of ff size 12{f} {} is given, so we must calculate net FnetFnet size 12{F} {}. That can be done since both the acceleration and mass of System 2 are known. Using Newton’s second law we see that

Fnet=ma,Fnet=ma size 12{F rSub { size 8{"net"} } = ital "ma"} {},

(8)where the mass of System 2 is 19.0 kg (mm size 12{m} {}= 12.0 kg + 7.0 kg) and its acceleration was found to be a=1.5 m/s2a=1.5 m/s2 size 12{a=1 "." "50"" m/s" rSup { size 8{2} } } {} in the previous example. Thus,

Fnet=ma,Fnet=ma size 12{F rSub { size 8{"net"} } = ital "ma"} {},

(9)
Fnet=(19.0 kg)(1.5 m/s2)=29 N.Fnet=(19.0 kg)(1.5 m/s2)=29 N size 12{F rSub { size 8{"net"} } = \( "19" "." "0 kg" \) \( 1 "." "50 m/s" rSup { size 8{2} } \) ="28" "." 5" N"} {}.

(10)Now we can find the desired force:

Fprof=Fnet+f,Fprof=Fnet+f size 12{F rSub { size 8{"prof"} } =F rSub { size 8{"net"} } +f} {},

(11)Fprof=29 N+24.0 N=53 N.Fprof=29 N+24.0 N=53 N size 12{F rSub { size 8{"prof"} } ="28" "." 5" N "+"24" "." "0 N "="52" "." "5 N"} {}.

(12)
*Discussion*

It is interesting that this force is significantly less than the 150-N force the professor exerted backward on the floor. Not all of that 150-N force is transmitted to the cart; some of it accelerates the professor.

The choice of a system is an important analytical step both in solving problems and in thoroughly understanding the physics of the situation (which is not necessarily the same thing).

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