A20 kg sled carrying a 40 kg boy is sliding at 12 m/s on smooth, level ice, when it encounters a rough patch of snow. a) what is the initial kinetic energy of the boy and the sled? b) if the rough ice exerts an average opposing force of 540 n, in what distance does the sled stop? c) what work is done by the rough ice stopping the sled? ( show work! )

a) 4,320 J

b) 8 m

c) 4,320 J

Explanation:

1) Data:

Part a) smooth ice:

m₁ = 20 kgm₂ = 40 kgv = 12 m/sKE = ?

Part b) rough ice:

F = 540 Nstopping distace, d = ?

Part c) rough ice:

W = ?

2) Solution:

a) KInetic energy:

KE = (1/2)mv² = (1/2) [m₁ + m₂] v² = (1/2) [20kg + 40kg] (12 m/s)² = 4,320 J

b) Distance to stop (Vf = 0)

Since, you have average force (opposing force) you can calculate average acceleration (deceleration)

F = m×a ⇒ a = F/m = F / (m₁ + m₂) = 540 N / (20 + 40 kg) = 9 m/s²

Now, use kinematic equation for uniformly decelarated motion:

Vf²= Vo² - 2ad ⇒ d = [Vo² - Vf²] /(2a) = [ (12m/s)² - 0 ] / (2×9m/s²) = 8 m

c) Work, W

Again, using average force, you can calculate work done by the force:

W = F × d = 540N × 8 m = 4,320 J

Note that this result confirms our calculations are fine, as the work done by the rough ice is equal to the initial kinetic energy, condition of energy conservation law.

The answer is a i think

Indoraptor for sure