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Using conservation of mechanical energy to calculate the speed of a toy car

A 0.100-kg toy car is propelled by a compressed spring, as shown in [link] . The car follows a track that rises 0.180 m above the starting point. The spring is compressed 4.00 cm and has a force constant of 250.0 N/m. Assuming work done by friction to be negligible, find (a) how fast the car is going before it starts up the slope and (b) how fast it is going at the top of the slope.

The figure shows a toy race car that has just been released from a spring. Two possible paths for the car are shown. One path has a gradual upward incline, leveling off at a height of eighteen centimeters above its starting level. An alternative path shows the car descending from its starting point, making a loop, and then ascending back up and leveling off at a height of eighteen centimeters above its starting level.
A toy car is pushed by a compressed spring and coasts up a slope. Assuming negligible friction, the potential energy in the spring is first completely converted to kinetic energy, and then to a combination of kinetic and gravitational potential energy as the car rises. The details of the path are unimportant because all forces are conservative—the car would have the same final speed if it took the alternate path shown.

Strategy

The spring force and the gravitational force are conservative forces, so conservation of mechanical energy can be used. Thus,

KE i + PE i = KE f + PE f size 12{"KE""" lSub { size 8{i} } +"PE" rSub { size 8{i} } ="KE" rSub { size 8{f} } +"PE" rSub { size 8{f} } } {}

or

1 2 mv i 2 + mgh i + 1 2 kx i 2 = 1 2 mv f 2 + mgh f + 1 2 kx f 2 ,

where h size 12{h} {} is the height (vertical position) and x size 12{x} {} is the compression of the spring. This general statement looks complex but becomes much simpler when we start considering specific situations. First, we must identify the initial and final conditions in a problem; then, we enter them into the last equation to solve for an unknown.

Solution for (a)

This part of the problem is limited to conditions just before the car is released and just after it leaves the spring. Take the initial height to be zero, so that both h i size 12{h rSub { size 8{i} } } {} and h f size 12{h rSub { size 8{f} } } {} are zero. Furthermore, the initial speed v i size 12{v rSub { size 8{i} } } {} is zero and the final compression of the spring x f size 12{x rSub { size 8{f} } } {} is zero, and so several terms in the conservation of mechanical energy equation are zero and it simplifies to

1 2 kx i 2 = 1 2 mv f 2 .

In other words, the initial potential energy in the spring is converted completely to kinetic energy in the absence of friction. Solving for the final speed and entering known values yields

v f = k m x i = 250 .0 N/m 0.100 kg ( 0.0400 m ) = 2.00 m/s. alignl { stack { size 12{v rSub { size 8{f} } = sqrt { { {k} over {m} } } x rSub { size 8{i} } } {} #" "= sqrt { { {"250" "." 0" N/m"} over {0 "." "100 kg"} } } \( 0 "." "0400"" m" \) {} # " "=2 "." "00"" m/s" "." {}} } {}

Solution for (b)

One method of finding the speed at the top of the slope is to consider conditions just before the car is released and just after it reaches the top of the slope, completely ignoring everything in between. Doing the same type of analysis to find which terms are zero, the conservation of mechanical energy becomes

1 2 kx i  2 = 1  2 mv f  2 + mgh f . size 12{ { {1} over {2} } ital "kx" rSub { size 8{i} rSup { size 8{2} } } = { {1} over {2} } ital "mv" rSub { size 8{f} rSup { size 8{2} } } + ital "mgh" rSub { size 8{f} } } {}

This form of the equation means that the spring’s initial potential energy is converted partly to gravitational potential energy and partly to kinetic energy. The final speed at the top of the slope will be less than at the bottom. Solving for v f size 12{v rSub { size 8{f} } } {} and substituting known values gives

v f = kx i 2 m 2 gh f = 250.0 N/m 0.100 kg ( 0.0400 m ) 2 2 ( 9.80 m/s 2 ) ( 0.180 m ) = 0.687 m/s. alignl { stack { size 12{v rSub { size 8{f} } = sqrt { { { ital "kx" rSub { size 8{i} rSup { size 8{2} } } } over {m} } - 2 ital "gh" rSub { size 8{f} } } } {} #" "= sqrt { left ( { {"250" "." 0" N/m"} over {0 "." "100 kg"} } right )"" \( 0 "." "0400"" m" \) rSup { size 8{2} } - 2 \( 9 "." "80"" m/s" rSup { size 8{2} } \) \( 0 "." "180"" m" \) } {} # " "=0 "." "687"" m/s" "." {}} } {}

Discussion

Another way to solve this problem is to realize that the car’s kinetic energy before it goes up the slope is converted partly to potential energy—that is, to take the final conditions in part (a) to be the initial conditions in part (b).

Section summary

  • A conservative force is one for which work depends only on the starting and ending points of a motion, not on the path taken.
  • We can define potential energy ( PE ) size 12{ \( "PE" \) } {} for any conservative force, just as we defined PE g size 12{"PE" rSub { size 8{g} } } {} for the gravitational force.
  • The potential energy of a spring is PE s = 1 2 kx 2 size 12{"PE" rSub { size 8{s} } = { {1} over {2} } ital "kx" rSup { size 8{2} } } {} , where k size 12{k} {} is the spring’s force constant and x size 12{x} {} is the displacement from its undeformed position.
  • Mechanical energy is defined to be KE + PE size 12{"KE "+" PE"} {} for a conservative force.
  • When only conservative forces act on and within a system, the total mechanical energy is constant. In equation form,
KE + PE = constant     or KE i + PE i = KE f + PE f (conservative forces only),

where i and f denote initial and final values. This is known as the conservation of mechanical energy.

Conceptual questions

What is a conservative force?

The force exerted by a diving board is conservative, provided the internal friction is negligible. Assuming friction is negligible, describe changes in the potential energy of a diving board as a swimmer dives from it, starting just before the swimmer steps on the board until just after his feet leave it.

Problems&Exercises

A 5 . 00 × 10 5 -kg size 12{5 "." "00" times "10" rSup { size 8{5} } "-kg"} {} subway train is brought to a stop from a speed of 0.500 m/s in 0.400 m by a large spring bumper at the end of its track. What is the force constant k size 12{k} {} of the spring?

7.81 × 10 5 N/m size 12{7 "." "81" times "10" rSup { size 8{5} } " N/m"} {}

A pogo stick has a spring with a force constant of 2 . 50 × 10 4 N/m size 12{2 "." "50" times "10" rSup { size 8{4} } " N/m"} {} , which can be compressed 12.0 cm. To what maximum height can a child jump on the stick using only the energy in the spring, if the child and stick have a total mass of 40.0 kg? Explicitly show how you follow the steps in the Problem-Solving Strategies for Energy .

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Source:  OpenStax, Unit 5 - work and energy. OpenStax CNX. Jan 02, 2016 Download for free at https://legacy.cnx.org/content/col11946/1.1
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