A 150.0 kg cart rides down a set of tracks on four solid steel wheels, each with radius 20.0 cm and mass 45.0 kg. The tracks slope downward at an angle of 17 ∘ to the horizontal. If the cart is released from rest a distance of 27.0 m from the bottom of the track (measured along the slope), how fast will it be moving when it reaches the bottom

Explanation:

given

Mass of cart= 150kg

mass of each wheel=45kg

mass of 4 wheels= 180kg

angle of the track=  17 ∘

distance of track= 27m

The height  of the tracl is calculated thus:

sin 17° = h / 27

h = sin 17*27

h=7.89m

"Potential energy at top = kinetic energy of cart and wheels at bottom + rotational energy of wheels at bottom "

1. Potential energy at top= (M+4m)gh

2. kinetic energy of cart and wheels at bottom= 1/2 (M+4m) v²

3. rotational energy of wheels at bottom= 4(1/2 Iω²)

The total is expressed as

(M+4m)gh = 1/2 (M+4m) v² + 4(1/2 Iω²) 1

we know that  I = mr² / 2

Put I= mr² / 2

(M+4m)gh = 1/2 (M+4m)v² + 4(1/2 (mr² / 2) ω²)

(M+4m)gh = 1/2 (M+4m)v² +  m r² ω²

we know that  v²= r² ω²

(M+4m)gh = 1/2 (M+4m)v² +  m v²

(M+4m)gh = v² (M/2 + 2m + m)

(M+4m)gh = v² (M/2 + 3m)

v = √[(M+4m)gh / (3m + M/2)]

v = √[(150 + 4*45) * 9.8 m/s² * 7.89 m / (3*45 + 150/2)]

v=√25516.26/142.5

v=√179.06

v = 13.4 m/s

a is the best answer your welcome