Question

a 40 kg skater starts from rest at point a along a frictionless ramp as shown below. the height of the ramp from the floor at the point b is yb= 3m and the speed of the skater at b is 5.0m/s

the mechanical energy of the skater just before hitting the ground at point c

Answer 1

By energy conservation,

Mechanical energy of the skater at point c (Tf) = Mechanical energy of the skater at point b (Ti)

Tf = Kinetic energy of skater at point b(KEi) + potential energy of skater at point b(PEi)

Tf = 0.5*m*v^2 + m*g*h

here, m = mass of skater = 40 kg

v = speed of skater at point b = 5.0 m/s

g = 9.81 m/s^2

h = height of point b = yb = 3 m

then, Tf = 0.5*40*5.0^2 + 40*9.81*3

Tf = 1677.2 J

In two significant figure:

Tf = 1.7*10^3 J

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