Figure one shows a metal block
attached to the end of a spring. Initially, the metal block is held
in place to counterbalance its weight, and the spring is in its unstretched
equilibrium position. The mass of the block is 15
kilograms. The gravitational field strength is
9.8 metres per second squared. Write down the equation that
relates the gravitational potential energy of an object to its height above the
ground, its mass, and the local gravitational field strength.
Since the equation we’re asked to
write is a general one, we won’t need to reference any of the specific information
about this particular system. We know that when it comes to
gravitational potential energy, this is given as the product of three factors. It’s equal to an object’s mass
multiplied by the local acceleration due to gravity multiplied by its height above
some reference point.
We often see this equation written
this way mathematically. And we can also express it using
words. We can say that gravitational
potential energy is equal to mass times acceleration due to gravity times
Now that we’ve written a general
equation for gravitational potential energy, let’s apply this relationship to our
The metal block is released from
its initial position. Calculate the change in
gravitational potential energy of the block from its initial position to when it is
10 centimetres below its initial position.
To solve for this change, we can
call the change in the height of the block given as 10 centimetres Δℎ. And we’ll also recall the mass of
the block given as 15 kilograms and the acceleration due to gravity 9.8 metres per
To calculate the block’s change in
gravitational potential energy, we can recall the equation we developed earlier. This equation tells us that the
gravitational potential energy of an object is equal to its mass times the
acceleration due to gravity times its height relative to some reference point.
In our case, we can say that the
change in gravitational potential energy is equal to 𝑚 times 𝑔 times Δℎ. All these values are given to
us. But before we plug them in, we want
to convert the units on our change in height from centimetres into units of
metres. This will help the units of that
value agree with the units in 𝑔.
To help with this conversion, we
can recall that 100 centimetres is equal to one metre. Therefore, 10 centimetres is equal
to 0.10 metres. We’re now ready to plug in and
solve for ΔPE sub 𝑔. When we do and we enter this
expression on our calculator, we find a result to three significant figures of 14.7
joules. That’s the change in the
gravitational potential energy of this block over this descent.
Next, let’s consider how the
potential energy of the block connects with its kinetic energy.
When the block was 10 centimetres
below its initial position, 58 percent of its initial gravitational potential energy
was transferred to its kinetic energy store. By how much did the block’s kinetic
energy increase after falling 10 centimetres? Give your answer to two significant
We can say that the kinetic energy
gained by the block is equal to the change in its gravitational potential energy
multiplied by 58 percent or as a decimal 0.58. We calculated earlier that ΔPE sub
𝑔 is 14.7 joules. This number times 0.58 is equal to
8.5 joules. To two significant figures, that’s
the kinetic energy gained by the block over this descent.
Next, let’s consider what this
kinetic energy means in terms of the motion of the block.
Calculate the speed of the block
after falling 10 centimetres. Give your answer to three
We saw earlier that over this
distance, the block’s change in kinetic energy was 8.5 joules. We can recall that the kinetic
energy of a classical object is equal to one-half its mass times its speed
squared. We can write then that ΔKE is equal
to one-half 𝑚𝑣 squared, where 𝑣 is the speed.
When we rearrange to solve for
speed by multiplying both sides by two over 𝑚 and taking the square root of both
sides, we find that the speed of the block is equal to the square root of two times
its change in kinetic energy divided by its mass.
We’ve calculated ΔKE and we know
its mass from an earlier step. So we’re ready to plug in and solve
for speed 𝑣. When we do plug in and enter this
expression on our calculator, to three significant figures, we find a result of 1.06
metres per second. That’s the speed of the block after
falling 10 centimetres.
Next, we’ll work with a type of
potential energy we haven’t considered so far — spring potential energy.
When the block reaches its lowest
point, the potential energy stored in the spring is 35.28 joules. Calculate the spring constant of
the spring. Use the correct equation from the
Physics Equation Sheet.
The potential energy stored in a
spring, which we can call PE sub 𝑠, is equal to one-half times the spring constant
𝑘 multiplied by the spring’s displacement from equilibrium 𝑥 squared. In our question statement, we’re
told the spring’s potential energy is 35.28 joules and we want to solve for 𝑘 based
on this information.
We can start by algebraically
rearranging this equation to solve for 𝑘. To do it, we’ll multiply both sides
by two divided by 𝑥 squared. The spring constant 𝑘 then is
equal to two times PE sub 𝑠 over 𝑥 squared.
We’re given the potential energy of
the spring and we’re told that 𝑥, this distance from equilibrium, occurs when the
block reaches its lowest point. When we consult Figure one, we see
that this lowest point occurs when the spring is extended by 24 centimetres. This tells us that 𝑥 is 24
But before we plug in this value
for 𝑥, we want to convert it into units of metres. Recalling that 100 centimetres is
equal to one metre, this means that 24 centimetres is equal to 0.24 metres. When we plug in this value for 𝑥
as well as the given value for PE sub 𝑠, to four significant figures, we find that
𝑘 is 1225 newtons per metre. That’s the spring constant of this