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Step Derivation/Formula Reasoning
1 [katex]h = s \sin(\theta)[/katex] To find the vertical height [katex]h[/katex] gained by the skier on the slope (where [katex]s = 695[/katex] meters, [katex]\theta = 34^\circ[/katex]), we use the sine component of the inclined angle because height is opposed to gravity and directly vertical.
2 [katex]h = 695 \sin(34^\circ)[/katex] Calculate [katex]h[/katex] by substituting the values of [katex]s[/katex] and [katex]\theta[/katex]. The sine of [katex]34^\circ[/katex] (from calculator or trigonometric table) is used to obtain [katex]h[/katex].
3 [katex]PE = mgh[/katex] Calculate the potential energy ([katex]PE[/katex]) needed to lift a skier to height [katex]h[/katex]. Here, [katex]m[/katex] is mass (72 kg), [katex]g[/katex] is the acceleration due to gravity (approximately [katex]9.81 \, \text{m/s}^2[/katex]), and [katex]h[/katex] is the height obtained in the previous step.
4 [katex]PE = 72 \cdot 9.81 \cdot 695 \sin(34^\circ)[/katex] Substitute the values into the potential energy formula to find the energy required to transport one skier to the top of the slope.
5 [katex]P = \frac{PE \cdot R}{t}[/katex] Calculate the power ([katex]P[/katex]) needed to transport [katex]R[/katex] riders per minute. [katex]t[/katex] is the time in seconds; for one minute, [katex]t = 60[/katex] seconds.
6 [katex]P = \frac{72 \cdot 9.81 \cdot 695 \sin(34^\circ) \cdot 5}{60}[/katex] Substitute the values to find the power required for ferrying 5 riders per minute.
7 [katex]P_{\text{total}} = \frac{P}{\text{efficiency}}[/katex] To find the total average power ([katex]P_{\text{total}}[/katex]) supplied by the motor, we adjust for the efficiency of the ski lift, which only uses 65% of the energy supplied to overcome work against friction.
8 [katex]P_{\text{total}} = \frac{72 \cdot 9.81 \cdot 695 \sin(34^\circ) \cdot 5}{60 \cdot 0.65}[/katex] Calculate [katex]P_{\text{total}}[/katex] by considering the efficiency. This power value indicates the power that must be supplied by the motor.
9 [katex]P_{\text{total}} \approx 35,192.76 \, \text{Watt}[/katex] Complete the calculations to get the final answer in Watts (rounded to sensible precision).

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35,192.76 Watts

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

General Metric Conversion Chart

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

1. Start with the given measurement: [katex]\text{5 km}[/katex]

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

3. Perform the multiplication: [katex]\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}[/katex]

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

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

[katex]10^{-12}[/katex]

Nano-

n

[katex]10^{-9}[/katex]

Micro-

µ

[katex]10^{-6}[/katex]

Milli-

m

[katex]10^{-3}[/katex]

Centi-

c

[katex]10^{-2}[/katex]

Deci-

d

[katex]10^{-1}[/katex]

(Base unit)

[katex]10^{0}[/katex]

Deca- or Deka-

da

[katex]10^{1}[/katex]

Hecto-

h

[katex]10^{2}[/katex]

Kilo-

k

[katex]10^{3}[/katex]

Mega-

M

[katex]10^{6}[/katex]

Giga-

G

[katex]10^{9}[/katex]

Tera-

T

[katex]10^{12}[/katex]

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 [katex] v_i [/katex] is written as [katex] u [/katex]; sometimes [katex] \Delta x [/katex] is written as [katex] s [/katex].
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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