Long Slides and Friction

Answer: 7300(J)

1. Find the length of the slide’s slope

Let $h$h be the height of the slide above the ground and $\theta$θ be its angle of inclination. The length $L$L of the slide’s slope is given by the following equation:

$L=\dfrac{h}{\sin \theta}$L=sinθh​

Substitute the given values into the equation.

$L=\dfrac{50}{0.342}\ \mathrm{m}$L=0.34250​ m

(As an aside, that’s roughly 146.198830409 m. That’s a long slide!)

2. Calculate the kinetic friction force

The normal force $N$N acting on Person A on the slope can be expressed as follows, where $m_0$m0​ is the mass and $g$g is the acceleration due to gravity:

$N=m_0g\cos\theta$N=m0​gcosθ

Let $\mu$μ be the coefficient of kinetic friction between the cardboard and the roller. The kinetic friction force $F_f$Ff​ acting during sliding is given by:

$F_f=\mu N=\mu m_0g\cos\theta$Ff​=μN=μm0​gcosθ

Substitute the given values into the equation.

$F_f = 0.08 \times 68 \times 9.8 \times 0.94\ \mathrm{N}$Ff​=0.08×68×9.8×0.94 N

3. Calculate the total frictional heat

The total frictional heat $Q$Q (the amount of work done by the kinetic friction force) is found by multiplying the kinetic friction force $F_f$Ff​ by the distance traveled $L$L.

$Q=F_f\times L=(\mu m_0g\cos\theta)\times(\dfrac{h}{\sin \theta})=\mu m_0gh\dfrac{\cos\theta}{\sin\theta}$Q=Ff​×L=(μm0​gcosθ)×(sinθh​)=μm0​ghsinθcosθ​

Substitute the values from the constraints directly into the equation.

$Q = 0.08 \times 68 \times 9.8 \times 50 \times \dfrac{0.94}{0.342}$Q=0.08×68×9.8×50×0.3420.94​

$Q = 2665.6 \times 2.748538 \approx 7326.50\ \mathrm{J}$Q=2665.6×2.748538≈7326.50 J

###4. Handling Significant Figures Following the instructions in the problem, the calculation yields “7300.”

The total frictional heat generated while Person A slid down is $\boxed{7300\ \mathrm{J}}$7300 J​.






It looks like we’ll be able to warm the water up a little!!