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Force on the cart—choosing a new system

Calculate the force the professor exerts on the cart in [link] 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 [link] ), 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, F prof size 12{F rSub { size 8{"prof"} } } {} , is an external force acting on System 2. F prof 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 F prof size 12{F rSub { size 8{"prof"} } } {} . Starting with

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

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

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

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

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

The value of f size 12{f} {} is given, so we must calculate net F net 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

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

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

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

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

Now we can find the desired force:

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

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

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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Phet explorations: gravity force lab

Visualize the gravitational force that two objects exert on each other. Change properties of the objects in order to see how it changes the gravity force.

Gravity Force Lab

Test prep for ap courses

What object or objects commonly exert forces on the following objects in motion? (a) a soccer ball being kicked, (b) a dolphin jumping, (c) a parachutist drifting to Earth.

(a) A soccer player, gravity, air, and friction commonly exert forces on a soccer ball being kicked.

(b) Gravity and the surrounding water commonly exert forces on a dolphin jumping. (The dolphin moves its muscles to exert a force on the water. The water exerts an equal force on the dolphin, resulting in the dolphin’s motion.)

(c) Gravity and air exert forces on a parachutist drifting to Earth.

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A ball with a mass of 0.25 kg hits a gym ceiling with a force of 78.0 N. What happens next?

  1. The ball accelerates downward with a force of 80.5 N.
  2. The ball accelerates downward with a force of 78.0 N.
  3. The ball accelerates downward with a force of 2.45 N.
  4. It depends on the height of the ceiling.
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Which of the following is true?

  1. Earth exerts a force due to gravity on your body, and your body exerts a smaller force on the Earth, because your mass is smaller than the mass of the Earth.
  2. The Moon orbits the Earth because the Earth exerts a force on the Moon and the Moon exerts a force equal in magnitude and direction on the Earth.
  3. A rocket taking off exerts a force on the Earth equal to the force the Earth exerts on the rocket.
  4. An airplane cruising at a constant speed is not affected by gravity.

(c)

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Practice Key Terms 2

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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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