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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

Respuesta :

hamzaahmeds

Answer:

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

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