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a student who weighs 556 Newtons climbs the stairway (vertical height of 4.0 m) in 25 s (a) how much work is done (b) what is the power output of the student

Sagot :

Given data:

Weight of the student;

[tex]\begin{gathered} F=mg \\ =556\text{ N} \end{gathered}[/tex]

Height;

[tex]h=4.0\text{ m}[/tex]

Time;

[tex]t=25\text{ s}[/tex]

Part (a)

The work done by the student is given as;

[tex]W=Fh[/tex]

Substituting all known values,

[tex]\begin{gathered} W=(556\text{ N})\times(4.0\text{ m}) \\ =2224\text{ J} \end{gathered}[/tex]

Therefore, the work done by the student is 2224 J.

Part (b)

The power is defined as the rate of doing work. Mathematically,

[tex]P=\frac{W}{t}[/tex]

Substituting alll known values,

[tex]\begin{gathered} P=\frac{2224\text{ J}}{25\text{ s}} \\ =88.96\text{ W} \end{gathered}[/tex]

Therefore, the power output of the student is 88.96 W.