A 0.100-kg toy car is propelled by a compressed spring, as shown in Figure 3. 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.

*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
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} } } {}

(8)or

12
mvi2
+
mgh
i
+12
kxi2
=12
mvf2+
mgh
f
+
12
kxf2,
12
mvi2
+
mgh
i
+12
kxi2
=12
mvf2+
mgh
f
+
12
kxf2,

(9)where hh size 12{h} {} is the height (vertical position) and xx 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 hihi size 12{h rSub { size 8{i} } } {} and hfhf size 12{h rSub { size 8{f} } } {} are zero. Furthermore, the initial speed vivi size 12{v rSub { size 8{i} } } {} is zero and the final compression of the spring xfxf 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
.
1
2
kx
i
2
=
1
2
mv
f
2
.

(10)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.
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" "." {}
} } {}

(11)*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

12
kxi 2
=1 2
mvf 2
+mghf.12
kxi 2
=1 2
mvf 2
+mghf. 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} } } {}

(12)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 vfvf 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.
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" "." {}
} } {}

(13)*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).

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