Question

The spring shown in (Figure 1) is compressed 42 cm and used to launch a 100 kg physics student. The track is frictionless until it starts up the incline. The student's coefficient of kinetic friction on the 30 degree incline is 0.16 .

How far up the incline does the student go? (I calculated v to be 12m/s)

Answer 1

Answer:

26.9 m

Explanation:

Method 1: Use work-energy theorem.

Draw a free body diagram of the student on the incline. There are 3 forces:

Weight force mg pulling down,

Normal force N pushing up perpendicular to the incline,

Friction force Nμ pushing down parallel to the incline.

Sum of forces in the perpendicular direction:

∑F = ma

N − mg cos θ = 0

N = mg cos θ

Therefore, friction force is Nμ = (mg cos θ) μ.

Initial elastic energy + initial potential = final potential energy + work done by friction

EE + PE₀ = PE + W

½ kx² + mgh₀ = mgh + Nμd

½ kx² + mgh₀ = mg (d sin θ) + (mg cos θ) μd

½ kx² + mgh₀ = mgd (sin θ + μ cos θ)

d = (½ kx² + mgh₀) / (mg (sin θ + μ cos θ))

Plug in values:

d = (½ (80,000) (0.42)² + (100) (9.8) (10)) / ((100) (9.8) (sin 30° + 0.16 cos 30°))

d = 26.9 m

Method 2: use force balance to find acceleration, then kinematics to find displacement.

First find the velocity at the bottom of the incline:

EE + PE₀ = KE

½ kx² + mgh₀ = ½ mv²

½ (80,000) (0.42)² + (100) (9.8) (10) = ½ (100) v²

v = 18.4 m/s

From the free body diagram, sum of forces in the parallel direction:

∑F = ma

-Nμ − mg sin θ = ma

-mgμ cos θ − mg sin θ = ma

-g (μ cos θ + sin θ) = a

-9.8 (0.16 cos 30° + sin 30°) = a

a = -6.26 m/s²

Given:

u = 18.4 m/s

v = 0 m/s

a = -6.26 m/s²

Find: s

v² = u² + 2as

0² = (18.4)² + 2 (-6.26) s

s = 26.9 m

Answer 2

Answer:

12 meters

Explanation:

The given equation represents the conservation of mechanical energy in this system, where the initial potential energy stored in the compressed spring is converted into kinetic energy as the student moves up the incline, and finally into gravitational potential energy at the highest point reached.

Let's break down the equation and solve for the distance $l$:

$\dfrac{1}{2} k x^2 + mgh = \mu mgl \cos(\theta) + mgxl \sin(\theta)$

where:

• $k$ = spring constant
• $x$ = compression of the spring
• $m$ = mass of the student
• $g$ = acceleration due to gravity
• $h$ = height above the reference point (initial height of the student)
• $\mu$ = coefficient of kinetic friction
• $l$ = distance up the incline
• $\theta$ = angle of the incline

Given values:

• $x = 0.42$ m (since the spring is compressed by 42 cm)
• $m = 100$ kg
• $g = 9.81 \, \textsf{m/s}^2$
• $\mu = 0.16$
• $\theta = 30^\circ$

Substituting these values into the equation and solving for $l$, we get:

$\begin{aligned} l &= \dfrac{\dfrac{1}{2} k x^2 + mgh}{mg[\mu \cos(\theta) + \sin(\theta)]} \\\\ &= \dfrac{\left(\dfrac{1}{2} \times 80000 \times (0.42)^2 + 100 \times 9.81 \times 0.42\right)}{100 \times 9.81 \times [0.16 \cos(30^\circ) + \sin(30^\circ)]} \\\\&\approx \dfrac{(0.5 \times 80000 \times 0.1764 + 100 \times 9.81 \times 0.42)}{100 \times 9.81 \times (0.16 \times 0.8660254037844 + 0.5)} \\\\&\approx \dfrac{(7056 + 412.02)}{100 \times 9.81 \times (0.6385640646)} \\\\&\approx \dfrac{7468.02}{626.4313474} \\\\ &\approx 11.92152984 \, \textsf{m} \\\\& \approx 12 \textsf{ m (in nearest whole number)} \end{aligned}$

So, the student goes approximately 12 meters up the incline.