Answer:
9.83 J
Explanation:
The force of friction causes the ball to decelerate as it slides. It also exerts a torque on the ball that causes it to rotationally accelerate while it slides. Eventually, the ball reaches a speed such that it rolls without sliding. To find the work done by friction at this point, we will need to use sum of forces, sum of torques, and conservation of energy.
First, the sum of forces on the ball is equal to its mass times acceleration.
∑F = ma
-mgμ = ma
The acceleration is equal to the change in velocity over time.
-mgμ = m (v − u) / t
Next, the sum of the torques about the ball's center is equal to its moment of inertia times its angular acceleration.
∑τ = Iα
The torque is equal to the force of friction times the radius, and the moment of inertia of a sphere is ⅖ mr².
mgμ r = ⅖ mr² α
mgμ = ⅖ mr α
The angular acceleration is equal to the change in angular velocity over time.
mgμ = ⅖ mr (ω − ω₀) / t
The initial angular velocity is 0.
mgμ = ⅖ mr ω / t
At the time that the ball begins to roll without slipping, v = ωr.
mgμ = ⅖ mv / t
Add the equations together and simplify.
0 = ⅖ mv / t + m (v − u) / t
0 = ⅖ v + v − u
u = ⁷/₅ v
v = ⁵/₇ u
Finally, energy is conserved, so the initial translational kinetic energy is equal to the final translational kinetic energy plus the final rotational kinetic energy plus the work done by friction.
KE₀ = KE + RE + W
½ mu² = ½ mv² + ½ Iω² + W
½ mu² = ½ mv² + ½ (⅖ mr²) (v/r)² + W
½ mu² = ½ mv² + ⅕ mv² + W
½ mu² = ⁷/₁₀ mv² + W
W = ½ mu² − ⁷/₁₀ mv²
Substitute the expression for final velocity found earlier:
W = ½ mu² − ⁷/₁₀ m (⁵/₇ u)²
W = ½ mu² − ⁵/₁₄ mu²
W = ¹/₇ mu²
Plug in values:
W = ¹/₇ (5.95 kg) (3.40 m/s)²
W = 9.83 J
The work done by friction before the ball begins to roll without slipping is 9.83 Joules. (Notice that the solution is independent of the coefficient of friction and the acceleration due to gravity.)
