Math  /  Algebra

QuestionFIGURE 5-38 Problem 26. (II) A child slides down a slide with a 2828^{\circ} incline, and at the bottom her speed is precisely half what it would have been if the slide had been frictionless. Calculate the coefficient of kinetic friction between the slide and the child.

Studdy Solution

STEP 1

1. The slide has an incline angle of 2828^{\circ}.
2. The child's speed at the bottom is half of what it would be if there were no friction.
3. We need to calculate the coefficient of kinetic friction (μk\mu_k) between the slide and the child.

STEP 2

1. Analyze the energy conservation for the frictionless slide.
2. Analyze the energy conservation for the slide with friction.
3. Relate the two scenarios to find the coefficient of kinetic friction.

STEP 3

Analyze the energy conservation for the frictionless slide.
- In the frictionless scenario, all potential energy converts to kinetic energy. - Let h h be the height of the slide, m m the mass of the child, and vf v_f the final speed without friction. - Potential energy at the top: mgh mgh . - Kinetic energy at the bottom: 12mvf2 \frac{1}{2}mv_f^2 . - By conservation of energy: mgh=12mvf2 mgh = \frac{1}{2}mv_f^2 .

STEP 4

Analyze the energy conservation for the slide with friction.
- In the scenario with friction, some energy is lost to friction. - Let v v be the final speed with friction. - Given that v=12vf v = \frac{1}{2}v_f . - Energy equation: mgh=12mv2+work done by friction mgh = \frac{1}{2}mv^2 + \text{work done by friction} . - Work done by friction: fkd=μkmgcos(θ)d f_k \cdot d = \mu_k \cdot mg \cdot \cos(\theta) \cdot d , where d d is the slide length.

STEP 5

Relate the two scenarios to find the coefficient of kinetic friction.
- From STEP_1, vf2=2gh v_f^2 = 2gh . - From STEP_2, v2=14vf2=14(2gh)=12gh v^2 = \frac{1}{4}v_f^2 = \frac{1}{4}(2gh) = \frac{1}{2}gh . - Substitute into energy equation: mgh=12m(12gh)+μkmgcos(θ)d mgh = \frac{1}{2}m(\frac{1}{2}gh) + \mu_k \cdot mg \cdot \cos(\theta) \cdot d . - Simplify: mgh14mgh=μkmgcos(θ)d mgh - \frac{1}{4}mgh = \mu_k \cdot mg \cdot \cos(\theta) \cdot d . - 34mgh=μkmgcos(θ)d \frac{3}{4}mgh = \mu_k \cdot mg \cdot \cos(\theta) \cdot d . - Cancel mg mg and solve for μk\mu_k: μk=3h4dcos(θ) \mu_k = \frac{3h}{4d \cos(\theta)} .

STEP 6

Express h h in terms of d d and θ\theta.
- From geometry, h=dsin(θ) h = d \cdot \sin(\theta) . - Substitute: μk=3dsin(θ)4dcos(θ) \mu_k = \frac{3d \cdot \sin(\theta)}{4d \cdot \cos(\theta)} . - Simplify: μk=3tan(θ)4 \mu_k = \frac{3 \tan(\theta)}{4} .

STEP 7

Calculate μk\mu_k using θ=28\theta = 28^{\circ}.
- Calculate tan(28)\tan(28^{\circ}). - μk=3tan(28)4\mu_k = \frac{3 \tan(28^{\circ})}{4}. - Use a calculator to find tan(28)0.5317\tan(28^{\circ}) \approx 0.5317. - μk=3×0.531740.3988\mu_k = \frac{3 \times 0.5317}{4} \approx 0.3988.
The coefficient of kinetic friction is approximately:
0.3988 \boxed{0.3988}

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