To solve this problem, we need to analyze the forces acting on the block as it slides down the incline in both scenarios.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( \text{E}_i \) is the initial mechanical energy | This energy remains constant in a frictionless environment without any external work done on the system. |
| 2 | Forces exerted: Horizontal \( F_H \) and Vertical \( F_V \) | When forces \( F_H \) and \( F_V \) are combined the net force is perpendicular to the ramp. Therefore this force does NOT do work on the block because the NET force act perpendicular to the motion. |
| 3 | \( W = \vec{F} \cdot \Delta \vec{x} \cdot \cos(\theta) \) | Work done \( W \) by a force is the product of the parallel force and displacement vectors. Here, \( \theta = 90^\circ \) for both forces. |
| 4 | \( \cos(90^\circ) = 0 \Rightarrow W = 0 \) | The forces do no work as their direction is perpendicular to the direction of motion. |
| 5 | \( \text{Mechanical Energy} = \text{E}_i \) | Since no work is done, there is no transformation of energy, and the mechanical energy of the system remains the same. |
| 6 | \text{(b) is equal to \( \text{E}_i \)}} | The correct answer is (b) because the additional forces do not alter the mechanical energy of the block-Earth system. |
Explanation of Incorrect Answers:**
– (a) Greater than \( \text{E}_i \): This cannot be true because the external forces do no work on the system.
– (c) Less than \( \text{E}_i \): This is also incorrect because the forces do not dissipate energy.
– (d) Cannot be determined without knowing \( F \): The magnitude of \( F \) does not affect the energy since no work is done.
Therefore, the mechanical energy remains unchanged at \( \text{E}_i \).
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A box of mass \(m\) is initially at rest at the top of a ramp that is at an angle \(\theta\) with the horizontal. The block is at a height \(h\) and length \(L\) from the bottom of the ramp. The coefficient of kinetic friction between the block and the ramp is \(\mu\). What is the kinetic energy of the box at the bottom of the ramp?
One end of a spring is attached to a solid wall while the other end just reaches to the edge of a horizontal, frictionless tabletop, which is a distance \(h\) above the floor. A block of mass \(M\) is placed against the end of the spring and pushed toward the wall until the spring has been compressed a distance \(x\). The block is released and strikes the floor a horizontal distance \(D\) from the edge of the table. Air resistance is negligible. Derive expressions for the following quantities only in terms of \(M, x, D, h,\) and any constants.
A sphere starts from rest and rolls down an incline of height \( H = 1.0 \) \( \text{m} \) at an angle of \( 25^\circ \) with the horizontal, as shown above. The radius of the sphere \( R = 15 \) \( \text{cm} \), and its mass \( m = 1.0 \) \( \text{kg} \). The moment of inertia for a sphere is \( \frac{2}{5}mR^2 \). What is the speed of the sphere when it reaches the bottom of the plane?
A constant force of strength \( 20 \) \( \text{N} \) acts on an object of mass \( 3 \) \( \text{kg} \) as it moves a distance of \( 4 \) \( \text{m} \). If this force is applied perpendicular to the \( 4 \) \( \text{m} \) displacement, the work done by the force is equal to:
A horizontal force of \(110 \, \text{N}\) is applied to a \(12 \, \text{kg}\) object, moving it \(6 \, \text{m}\) on a horizontal surface where the kinetic friction coefficient is \(\mu_k = 0.25\). The object then slides up a \(17^\circ\) inclined plane. Assuming the \(110 \, \text{N}\) force is no longer acting on the incline, and the coefficient of kinetic friction there is \(\mu_k = 0.45\), calculate the distance the object will slide on the incline.
A pendulum consists of a mass \( M \) hanging at the bottom end of a massless rod of length \( \ell \) which has a frictionless pivot at its top end. A mass \( m \), moving with velocity \( v \), impacts \( M \) and becomes embedded. In terms of the given variables and constants, what is the smallest value of \( v \) sufficient to cause the pendulum (with embedded mass \( m \)) to swing clear over the top of its arc?
A block is attached to a horizontal spring and is initially at rest at the equilibrium position \( x = 0 \), as shown in Figure \( 1 \). The block is then moved to position \( x = -A \), as shown in Figure \( 2 \), and released from rest, undergoing simple harmonic motion. At the instant the block reaches position \( x = +A \), another identical block is dropped onto and sticks to the block, as shown in Figure \( 3 \). The two–block–spring system then continues to undergo simple harmonic motion. Which of the following correctly compares the total mechanical energy \( E_{\text{tot},2} \) of the two–block–spring system after the collision to the total mechanical energy \( E_{\text{tot},1} \) of the one–block–spring system before the collision?
A spring in a pogo-stick is compressed \( 12 \) \( \text{cm} \) when a \( 40. \) \( \text{kg} \) girl stands on it. What is the spring constant for the pogo-stick spring?
A crate is pulled 2.5 m at constant velocity along a 25° incline. The coefficient of kinetic friction between the crate and the plane is 0.250. What is the efficiency of this procedure?
In the figure above, the marble rolls down the track and around a loop-the-loop of radius \( R \). The marble has mass \( m \) and radius \( r \). What minimum height \( h_{min} \) must the track have for the marble to make it around the loop-the-loop without falling off? Express your answer in terms of the variables \( R \) and \( r \).
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| \(W = Fd \cos\theta\) |
| Momentum | Torque and Rotations |
|---|---|
| \(p = mv\) | \(\tau = r \cdot F \cdot \sin(\theta)\) |
| \(J = \Delta p\) | \(I = \sum mr^2\) |
| \(p_i = p_f\) | \(L = I \cdot \omega\) |
| Simple Harmonic Motion | Fluids |
|---|---|
| \(F = -kx\) | \(P = \frac{F}{A}\) |
| \(T = 2\pi \sqrt{\frac{l}{g}}\) | \(P_{\text{total}} = P_{\text{atm}} + \rho gh\) |
| \(T = 2\pi \sqrt{\frac{m}{k}}\) | \(Q = Av\) |
| \(x(t) = A \cos(\omega t + \phi)\) | \(F_b = \rho V g\) |
| \(a = -\omega^2 x\) | \(A_1v_1 = A_2v_2\) |
| Constant | Description |
|---|---|
| [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 |
| Variable | SI 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] |
| Variable | Derived 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]\omega[/katex] (Angular Velocity) | [katex]\text{radians per second (rad/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] |
Metric Prefixes
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: [katex]\text{5 km}[/katex]
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]
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]
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] |
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