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alukav5142 [94]
3 years ago
14

A 66.5-kg hiker starts at an elevation of 1270 m and climbs to the top of a peak 2660 m high.

Physics
1 answer:
anyanavicka [17]3 years ago
8 0

Answer

given,

mass of the hiker = 66.5 kg

elevation of the hiker,h₁ = 1270 m

elevation of top peak,h₂ = 2660 m

a) Change in potential energy

    Δ PE = m g h₂ - m g h₁

    Δ PE = m g (h₂ - h₁)

    Δ PE = 66.5 x 9.8 x  (2660 - 1270)

    Δ PE = 905863 J

b) Minimum work require by the hiker will be equal to Δ PE = 905863 J

c) yes, it can be more than this, if friction is present on the surface.

   

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v_y^2 - u_y^2 = 2as

v_y = \sqrt{u^2+2as}=\sqrt{(-20.0)^2+2(-9.8)(-25)}=-29.8 m/s (downward, so we take the negative solution)

As before, the horizontal speed instead is zero, since the pebble is initially launched vertically, so the final speed is just equal to the magnitude of the vertical velocity:

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