Question

1) You and some friends are enjoying a trip to the water park on a hot summer day! Upon arrival, you decide you want to try the fastest (and scariest) water slide they have. To ride this water slide, a rider stands on a retractable floor and leans up against an almost vertical portion of water slide, that makes an angle of approximately 84° with the horizontal. The operatorpresses a button, the floor is retracted, and riders slide straight at that angle for approximately 6.3 meters until they are dropped into the twisting turning water slide. You (m= 62 kg) start at the top from rest. If the coefficient of kinetic friction between you and the slide is 0.12, how fast are you moving at the end of the straight portion of the slide? (Hint: set the y=0 level tothe bottom of the straight portion of the slide!)

Answer 1

ANSWER

11.02 m/s

EXPLANATION

Given:

• Your mass, m = 62 kg

,

• Your initial velocity, u = 0 (you start from rest at the top)

,

• The coefficient of kinetic friction between you and the slide, μk = 0.12

,

• The angle between the slide and the horizontal, θ = 84°

,

• The length of the slide, d = 6.3 m

Let's make a diagram of the description of the slide given,

First, we can find what is your gravitational potential energy just before the floor is retracted,

$PE=mgh$

We can find the vertical height of the slide using the right triangle formed,

$\sin\theta=\frac{h}{d}\Rightarrow h=d\sin\theta=6.3m\cdot\sin84\degree\approx6.27m$

So your gravitational potential energy is,

$PE=62kg\cdot\cdot9.8m/s^2\cdot6.27m=3809.65J$

If there was no friction between you and the slide, this energy would be conserved and it would be equal to your kinetic energy when you reach the bottom of the straight portion of the slide. However, there is indeed friction, so some of this energy will be lost to the work done by the kinetic frictional force.

To find the mentioned work, we have to draw a force diagram of you sliding down the slide and find the kinetic frictional force,

By Newton's second law,

$F_N-F_{gy}=0$

Using trigonometry, we have that the weight in the y-direction is,

$F_N=F_{gy}=F_g\cos\theta=mg\cos\theta$

So, the frictional force is,

$F_f=\mu_k\cdot F_N=\mu_k\cdot m\cdot g\cdot\cos\theta=0.12\cdot62kg\cdot9.8m/s^2\cdot\cos84\degree\approx7.62N$

And this force acts for a distance of 6.3 meters, so the work it does on the person sliding down the slide is,

$W=F_f\cdot d=7.62N\cdot6.3m=48.01J$

This is the energy lost to friction, so the kinetic energy you will have at the end of the straight portion of the slide is,

$KE=PE-W=3809.65J-48.01J=3761.64J$

This kinetic energy is equal to,

$KE=\frac{1}{2}mv^2$

Solving for v,

$v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2\cdot3761.64J}{62kg}}\approx11.02m/s$

Hence, your speed at the end of the straight portion of the slide is 11.02 m/s, rounded to two decimal places.