# A spring launches a 4 kg block across a frictionless horizontal surface. The block then ascends a 30° incline with a kinetic friction coefficient of 0.25, stopping after 55 m on the incline. If the spring constant is 800 N/m, find the initial compression of the spring. Disregard friction while in contact with the spring.

x = \boxed{1.965\, m}
0
Step Derivation/Formula Reasoning
1 U_{\text{spring}} = \frac{1}{2} k x^2 The potential energy stored in the compressed spring, where k is the spring constant and x is the compression distance.
2 W_{\text{friction}} = \mu_k \cdot m \cdot g \cdot \cos(\theta) \cdot d Work done against friction while moving up the incline: \mu_k is the coefficient of kinetic friction, m is mass, g is acceleration due to gravity, \theta is the angle of incline, and d is the distance traveled up the incline.
3 PE = m \cdot g \cdot h The gravitational potential energy gained by the block at the end of the incline. Here, h is the vertical height, calculated further in Step 5.
4 U_{\text{spring}} = W_{\text{friction}} + PE Conservation of mechanical energy: The total mechanical energy (spring potential energy) initially equals the sum of the work done against friction and the gravitational potential energy at the end.
5 h = d \cdot \sin(\theta) Using the distance d traveled up the incline and the angle \theta to calculate the height h .
6 Substituting h , W_{\text{friction}} , and PE in step 4: \frac{1}{2} k x^2 = \mu_k \cdot m \cdot g \cdot \cos(\theta) \cdot d + m \cdot g \cdot d \cdot \sin(\theta)
7 Simplifying and solving for x : x^2 = \frac{2(\mu_k \cdot m \cdot g \cdot \cos(\theta) \cdot d + m \cdot g \cdot d \cdot \sin(\theta))}{k}
x = \sqrt{\frac{2d(mg(\mu_k \cdot \cos(\theta) + \sin(\theta)))}{k}}
8 x = \sqrt{\frac{2 \cdot 55 \cdot (4 \cdot 9.8 \cdot (0.25 \cdot \cos(30^\circ) + \sin(30^\circ)))}{800}} Substitute values for d = 55 \, \text{m} , m = 4 \, \text{kg} , g = 9.8 \, \text{m/s}^2 , \mu_k = 0.25 , \theta = 30^\circ , and k = 800 \, \text{N/m} to find x .
9 x = \boxed{1.965\, m} Final value.

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x = \boxed{1.965\, m}

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KinematicsForces
\Delta x = v_i t + \frac{1}{2} at^2F = ma
v = v_i + atF_g = \frac{G m_1m_2}{r^2}
a = \frac{\Delta v}{\Delta t}f = \mu N
R = \frac{v_i^2 \sin(2\theta)}{g}
Circular MotionEnergy
F_c = \frac{mv^2}{r}KE = \frac{1}{2} mv^2
a_c = \frac{v^2}{r}PE = mgh
KE_i + PE_i = KE_f + PE_f
MomentumTorque and Rotations
p = m v\tau = r \cdot F \cdot \sin(\theta)
J = \Delta pI = \sum mr^2
p_i = p_fL = I \cdot \omega
Simple Harmonic Motion
F = -k x
T = 2\pi \sqrt{\frac{l}{g}}
T = 2\pi \sqrt{\frac{m}{k}}
ConstantDescription
gAcceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface
GUniversal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2
\mu_k and \mu_sCoefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion.
kSpring constant, in \text{N/m}
M_E = 5.972 \times 10^{24} , \text{kg} Mass of the Earth
M_M = 7.348 \times 10^{22} , \text{kg} Mass of the Moon
M_M = 1.989 \times 10^{30} , \text{kg} Mass of the Sun
VariableSI Unit
s (Displacement)\text{meters (m)}
v (Velocity)\text{meters per second (m/s)}
a (Acceleration)\text{meters per second squared (m/s}^2\text{)}
t (Time)\text{seconds (s)}
m (Mass)\text{kilograms (kg)}
VariableDerived SI Unit
F (Force)\text{newtons (N)}
E, PE, KE (Energy, Potential Energy, Kinetic Energy)\text{joules (J)}
P (Power)\text{watts (W)}
p (Momentum)\text{kilogram meters per second (kgm/s)}
\tau (Torque)\text{newton meters (Nm)}
I (Moment of Inertia)\text{kilogram meter squared (kgm}^2\text{)}
f (Frequency)\text{hertz (Hz)}

General Metric Conversion Chart

Example of using unit analysis: Convert 5 kilometers to millimeters.

1. Start with the given measurement: \text{5 km}

2. Use the conversion factors for kilometers to meters and meters to millimeters: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}

3. Perform the multiplication: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}

4. Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

10^{-12}

Nano-

n

10^{-9}

Micro-

µ

10^{-6}

Milli-

m

10^{-3}

Centi-

c

10^{-2}

Deci-

d

10^{-1}

(Base unit)

10^{0}

Deca- or Deka-

da

10^{1}

Hecto-

h

10^{2}

Kilo-

k

10^{3}

Mega-

M

10^{6}

Giga-

G

10^{9}

Tera-

T

10^{12}

1. Some answers may be slightly off by 1% depending on rounding, etc.
2. Answers will use different values of gravity. Some answers use 9.81 m/s2, and other 10 m/s2 for calculations.
3. Variables are sometimes written differently from class to class. For example, sometime initial velocity v_i is written as u ; sometimes \Delta x is written as s .
4. Bookmark questions that you can’t solve so you can come back to them later.
5. Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!

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