A wagon sits at the top of a
hill. The wagon is given a push that
increases its speed negligibly but is just sufficient to set the wagon in motion
down a straight slope. The wagon rolls 53.9 meters down
the slope, which is inclined 16.5 degrees below the horizontal, and reaches the
bottom of the hill. If friction is negligible, what
speed is the wagon moving when it reaches the bottom of the hill?
We can call the speed of the wagon
when it reaches the bottom of the hill 𝑣 sub 𝑓. And we’ll start on our solution by
drawing a diagram of the situation. A wagon sits perched at the top of
a long flat hill whose length we’ve called 𝐿. It’s 53.9 meters. The hill is inclined at an angle
below the horizontal — we’ve called 𝜃 — given as 16.5 degrees. We’re told the wagon is given a
very slight push to set it in motion and then rolls down the hill. And when it reaches the bottom of
the hill, we want to solve for its speed, 𝑣 sub 𝑓.
In this scenario, our system
consists of the wagon and the hill. And since energy is neither added
to nor taken away from this system, we can say that it’s conserved. Applying the general statement of
energy conservation to this scenario, we’ll use our freedom to decide when the
initial moment and when the final moments are. Choosing the initial moment to be
just when the cart is pushed at the top of the hill, and the final moment when it
just reaches the bottom of the hill.
We can expand this conservation
expression to say that the kinetic energy plus potential energy initially is equal
to the kinetic energy plus potential energy finally. At the outset, the wagon is not in
motion. So its initial kinetic energy is
zero. And if we choose the height at the
bottom of the hill to be our reference level of zero, we can say that the final
potential energy of the cart is also zero, since it’s at zero height. So the wagon’s initial potential
energy is equal to its final kinetic energy.
The wagon’s potential energy is
entirely gravitational. And we recall that gravitational
potential energy is equal to 𝑚 times 𝑔 times ℎ. Applying that relationship, we
write 𝑚 times 𝑔 times the initial height of the wagon equals KE sub 𝑓. We also recall that an object’s
kinetic energy equals half its mass times its speed squared. We can write this in for our KE sub
Looking at this equality, we see
that the mass of the wagon appears on both sides. So it cancels out. We want to solve for the final
speed of the wagon. So we rearrange this equation and
see the wagon’s final speed is equal to the square root of two times 𝑔 times ℎ sub
𝑖. The acceleration due to gravity,
𝑔, we take to be exactly 9.8 meters per second squared.
Looking at our diagram, we see that
the height of the hill, ℎ sub 𝑖, relates to the hill length, 𝐿, and the angle of
inclination, 𝜃. In particular, we can write that ℎ
sub 𝑖 is equal to 𝐿 times the sin of 𝜃. So we substitute 𝐿 sin 𝜃 in for ℎ
sub 𝑖 in our expression for 𝑣 sub 𝑓.
Looking at this expression, we’re
given the length 𝐿, the angle 𝜃, and 𝑔 is a known constant. So we’re ready to plug in and solve
for 𝑣 sub 𝑓. Entering these values on our
calculator, we find that, to three significant figures, 𝑣 sub 𝑓 is 17.3 meters per
second. That’s the speed of the wagon when
it reaches the bottom of the hill.