If the skater started from rest 4 m above the ground (instead of 7m), what would be the kinetic energy at the bottom of the ramp (which is still 1 m above the ground) ?

--> mass of skater is 75 kg and the acceleration of gravity is 9.81 N/kg

If the skater started from rest 4 above the ground (instead of 7), what would be the kinetic energy at the bottom of the ramp (which is still 1 above the ground) ?

a. 735 J

b. 2940 J

c. 2205 J

d. 4410 J

C. 2205 J

Explanation:

First we need to find the final velocity of skater at the bottom.

We, have:

Height lost = h = 4 m - 1 m = 3 m

Initial Velocity = Vi = 0 m/s (since, it starts from rest)

acceleration due to gravity = g = 9.8 m/s²

using third equation of motion:

2gh = Vf² - Vi²

2 (9.8 m/s²) (3 m) = Vf² - (0 m/s) ²

Vf² = 58.8 m²/s²

Now, for kinetic energy at bottom:

K. E = (1/2) m Vf²

K. E = (1/2) (75 kg) (58.8 m²/s²)

K. E = 2205 J

Hence, the correct option is C. 2205 J