Question

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

Answer 1

The final speed of the cart when it reaches the bottom of the track is 9.63 m/s.

Step-by-step explanation:

To find the final speed of the cart when it reaches the bottom of the track, we can use the principle of conservation of energy. Initially, the cart only has gravitational potential energy, which is converted into both kinetic energy and rotational kinetic energy when it reaches the bottom. The equation of conservation of energy is:

mgh = (1/2)mv^2 + (1/2)Iw^2

where m is the mass of the cart, g is the acceleration due to gravity, h is the vertical distance the cart traveled, v is the final speed of the cart, I is the moment of inertia of the wheels, and w is the final angular velocity of the wheels.

Since the wheels are solid and rolling without slipping, we can use the equation w = v/r, where r is the radius of the wheels. Substituting this into the conservation of energy equation and solving for v, we get:

v = sqrt(2g(h - I/(mr^2)))

Plugging in the given values, we have:

v = sqrt(2(9.8)(27 - (45/150)(0.2^2)))

v = 9.63 m/s

Answer 2

13.4 m/s

Step-by-step 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