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. A solid sphere of mass 4.0 kg and radius 0.20 m with a speed of 10 m/s rolls up from the

bottom of a ramp inclined 20˚. How high approximately the ball will stop on the incline from
the bottom?
a. 3.0 m b. 7.0 m c. 10 m d. 15 m e. 20 m

User Arcadio
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2 Answers

3 votes

Answer:

b. 7.0 m

Step-by-step explanation:

Energy is conserved.

Kinetic energy + rotational energy = potential energy

KE + RE = PE

½ mv² + ½ Iω² = mgh

Since the ball rolls without slipping, ω = v/r.

Moment of inertia for a solid sphere is I = ⅖ mr².

Substituting:

½ mv² + ½ (⅖ mr²) (v/r)² = mgh

½ mv² + ⅕ mv² = mgh

⁷/₁₀ v² = gh

h = ⁷/₁₀ v² / g

Plug in values:

h = ⁷/₁₀ (10 m/s)² / (10 m/s²)

h = 7.0 m

User Goowik
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4 votes

Final answer:

Using the conservation of mechanical energy and considering both translational and rotational kinetic energy of the rolling solid sphere, the height at which it will stop on the incline is approximately 7 meters. The correct answer is option: b. 7.0 m

Step-by-step explanation:

To determine how high the solid sphere will stop on the incline from the bottom, we apply the principle of conservation of mechanical energy, assuming no energy is lost due to friction or air resistance.

The kinetic energy of the sphere at the bottom of the ramp gets converted into potential energy as it climbs the ramp and comes to a stop.

The initial kinetic energy (KE) of the sphere includes both translational kinetic energy and rotational kinetic energy since the sphere is rolling. This is given by the sum of ½mv² (translational) and ½Iω² (rotational), where m is the mass, v is the velocity, I is the moment of inertia for a solid sphere (¾ m r²), and ω is the angular velocity.

The potential energy (PE) at the height h that the sphere will reach on the ramp is mgh, where g is the acceleration due to gravity.

At the point where the sphere stops, all the kinetic energy has been converted into potential energy.

Thus, mgh = ½mv² + ½Iω². Since the sphere rolls without slipping, v = rω. Substituting I and rearranging to solve for h, we get:

h = (7/10) (v²/g)

Using the provided values: v = 10 m/s, g = 9.8 m/s², and solving for h, we get:

h = (7/10) * (10² / 9.8)

≈ 7.14 m

Thus, the sphere will approximately stop 7 meters up the incline from the bottom, making option b the correct answer.

User JanHak
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