Atoboggan and rider have a total mass of 100 kg and travel down along the (smooth) slope defined by the equation y=0.2x2. at the instant x=8 m, the toboggan's speed is 4 m/s. at this point, determine the rate of increase in speed and the normal force which the toboggan exerts on the slope. neglect the size of the toboggan and rider for the calculation

9.35 m/s^2

n = 292.24

Explanation:

y = 0.2×(x^2) , the equation that desribes the slope is parabolic, then;

the tangent line at a point can be found by differentiating the equation:

dy/dx = 0.4×x

then, at x = 8 m:

dy/dx = 0.4×8 = 3.2

that represents the slope of the tangent line

then to find the incline:

tan(∅) = 3.2

      ∅ = 72.65°

at the point, the toboggan is sitting at 72.65° on the incline.

since the surface is smooth, no friction.

the only force playing is the parallel component of gravity, then by Newton's second law:

F = m×a

a = F/m

  = m×g×sin(∅)/m

  = g×sin(∅)

  = (9.8)×(sin(72.65°))

  = 9.35 m/s^2

then the normal force is given by:

n = m×g×cos(∅) = (100)×(9.8)×(cos(72.65))

n = 292.24

Use the surface of the water as a reference.
Measure distance, velocity, and acceleration as positive downward.

Define
t = time, s
h = distance, m
a = 10 m/s², acceleration
u = 16.5 ft/s, initial velocity

The velocity at time t is v = u + at
When the ball reaches the bottom of the lake at t = 5.7s, its velocity is
v = (16.5 ft/s) + (10 ft/s²)*(5.7 s) = 73.5 m/s

Distance traveled at time t is h = ut + (1/2)at².
The ball reaches the bottom of the lake at 5.7 s.
Therefore the depth of the lake is
H = (16.5 ft/s)*(5.7 s) + 0.5*(10 ft/s²)*(5.7 s)² = 256.5 m

The ball reaches the bottom of the lake with velocity 73.5 m/s
The depth of the lake is 256.5 m.