Question

A 55.9-kg hiker ascends a 46.6-meter high hill at a constant speed of 1.0 m/s. If it takes 373 s to climb the hill, then determine the power delivered by the hiker.

Answer 1

Step 1

State formula for Power

`\begin{gathered} \text{Power =}(work)/(\Delta time) \\ \text{But work = Force(F)}* Dis\tan ce(S)\text{ for rectilinear motion} \\ \end{gathered}`

Hence,

`\begin{gathered} \text{Power}=(F* S)/(\Delta Time(t)) \\ \text{Also Force(F) for rectilinear motion}=\text{Mass(M)}* acceleration(a) \\ \text{Hence,} \\ \text{Power}=\text{ }(M* a* S)/(\Delta t) \end{gathered}`

Step 2

Determine the power delivered by the hiker.

`\begin{gathered} \text{Power}=(M* a* S)/(\Delta t) \\ M=\text{mass = 55.9kg} \\ a=\text{ rectilinear acc}eleration\text{ = 1.0m/s} \\ S=\text{distance}=46.6m \\ \Delta t=373s \end{gathered}`
`\begin{gathered} \text{Power}=(55.9*1.0*46.6)/(373) \\ \end{gathered}`
`\begin{gathered} \text{Power}=6.983753351 \\ \text{Power }\approx\text{ 6.984 Joules(J)/Seconds(s) or Watts(w) to 3 decimal places} \\ \end{gathered}`

Hence the power delivered by the hiker to 3 decimal places is approximately 6.984 J/s or watts