A kid weighs 420 N, and she is sitting on a playground swing that hangs 0.40 m above the ground. Her mom pulls the swing back and releases it when the seat is 1.00 m above the ground. If she moves through the lowest point at 2.0 m/s, how much work was done on the swing by friction?

Approximately [tex](-166)\; {\rm J}[/tex], assuming that [tex]g = 9.81\; {\rm N \cdot kg^{-1}}[/tex] and that the mass of the playground swing is negligible.

Explanation:

If there is no energy loss, the gravitational potential energy released would be entirely converted into kinetic energy. However, because of friction, some of that energy is lost. The work that friction did on the swing would be equal to the difference between the following two:

Gravitational potential energy released when the height of the swing and the kid changed from [tex]1.00\; {\rm m}[/tex] to [tex]0.40\; {\rm m}[/tex].
Kinetic energy that the swing gained by the time it reached the lowest point.

To find the change in gravitational potential energy, multiply weight by the change in height:

[tex]\begin{aligned} & (\text{change in GPE}) \\ =\; & (\text{weight})\, (\text{change in height}) \\ =\; & (420\; {\rm N})\, (1.00\; {\rm m} - 0.40\; {\rm m}) \\ =\; & 252\; {\rm J}\end{aligned}[/tex].

In other words, [tex]252\; {\rm J}[/tex] of gravitational potential energy would be released when the swing traveled from [tex]1.00\; {\rm m}[/tex] to the lowest point at [tex]0.40\; {\rm m}[/tex].

If an object of mass [tex]m[/tex] is moving at a speed of [tex]v[/tex], the kinetic energy of that object would be:

[tex]\displaystyle (\text{KE}) = \frac{1}{2}\, m\, v^{2}[/tex].

In this question, while weight is given, mass is not. To find mass, divide weight by the gravitational field strength:

[tex]\begin{aligned}(\text{mass}) &= \frac{(\text{weight})}{g} \\ &= \frac{420\; {\rm N}}{9.81\; {\rm N\cdot kg^{-1}}} \\ &\approx 42.813\; {\rm kg}\end{aligned}[/tex].

At a speed of [tex]2.0\; {\rm m\cdot s^{-1}}[/tex], kinetic energy would be:

[tex]\begin{aligned} (\text{KE}) &= \frac{1}{2}\, m\, v^{2} \\ &\approx \frac{1}{2}\, (42.813\; {\rm kg}) \, (2.0\; {\rm m\cdot s^{-1}})^{2} \\ &\approx 85.627\;{\rm J}\end{aligned}[/tex].

Assuming that the kid and the swing were initially stationary with an initial velocity of [tex]0\; {\rm J}[/tex]. The change in kinetic energy would be approximately [tex]85.627\; {\rm J}[/tex].

Note that while [tex]252\; {\rm J}[/tex] of gravitational potential energy was released, kinetic energy increased by only [tex]85.627\; {\rm J}[/tex]. The difference between the two is the work of friction:

[tex]85.627\; {\rm J} - 252\; {\rm J} \approx (-166)\; {\rm J}[/tex].