A 68 kg hiker walks at 5.0 km/h up a 4% slope. The indicated incline is the ratio of the vertical distance and the horizontal distance expressed as percentage. What is the necessary metabolic power? Hint: You can model her power needs as the sum of the 380 W power to walk on level ground plus the power needed to raise her body by the appropriate amount. Assume that the efficiency of the body in using energy is 25%.

Explanation:

Power = work done/time

Where work = force × distance

Hence, power = force×distance/time1

Recall, velocity = distance/time

Hence equation one can be rewritten as

Power = force × velocity

5km/hr = 5×1000m/3600s

= 1.388m/s which is the velocity.

Force = Wy ×g

= Wsinθ ×g ( since we are considering the vertical component which is y ) 2

slope = tan θ

This indicates the ratio of the vertical and horizontal distance.

tan θ = 0.04

θ = 2.290˚

Therefore, substitute the values into equation 2

Force = 68 kg ×sin(2.290˚) × 9.8 m/s^2

F = 68×0.0399×9.8

Force = 26.627N

F = 26.63newton

Now, we shall calculate the power

Power = force× velocity

Power = 26.63×1.388

= 36.959 watts

It was stated that the hikers efficiency is 25%, to determine the metabolic power:

25% = 25/100

= 0.25

Metabolic power = 36.959/0.25

= 1478.36 watts

the simple machine has three main features. they are:

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever

frequency or wavelenth of em radiation in a vacuum