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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).

Applying the science practices: potential energy in a spring

Suppose you are running an experiment in which two 250 g carts connected by a spring (with spring constant 120 N/m) are run into a solid block, and the compression of the spring is measured. In one run of this experiment, the spring was measured to compress from its rest length of 5.0 cm to a minimum length of 2.0 cm. What was the potential energy stored in this system?

Answer

Note that the change in length of the spring is 3.0 cm. Hence we can apply Equation 7.42 to find that the potential energy is PE = (1/2)(120 N/m)(0.030 m) 2 = 0.0541 J.

Practice Key Terms 5

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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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