FE Civil Dynamics practice problems

Starting from rest with constant acceleration, the average velocity over the trip is simply half the final velocity. Distance equals average velocity times time, so $t = s / \bar{v} = 360 / 12 = 30\,\text{s}$. You can also find $a$ from the v-squared equation, then use $v = at$. The 15 s choice halves the time by using the final velocity (24) instead of the average velocity (12). Equivalently, it divides distance by the final velocity (360/24 = 15). The 7.5 s choice divides the distance by twice the final velocity. The 21.2 s choice comes from misusing $t = \sqrt{2s/a}$ with the wrong (doubled) acceleration.

1. From rest, find acceleration with the no-time equation:
   $v^2 = v_0^2 + 2a(s - s_0) \implies 24^2 = 0 + 2a(360)$
2. Solve for acceleration:
   $a = \frac{576}{720} = 0.8\,\text{m/s}^2$
3. Now find time from $v = v_0 + at$:
   $t = \frac{v - v_0}{a} = \frac{24 - 0}{0.8} = 30\,\text{s}$

FE Reference Handbook 10.6, p. 105

The height equation $y = v_0 t - \tfrac{1}{2}g t^2$ is quadratic in time, so it has two solutions for any height below the peak: one on the way up and one on the way down. You want the larger root (coming down). The 1.00 s choice is the smaller root (on the way up). The 6.12 s choice is the total time of flight back to the ground (2v_0/g). The 3.06 s choice is the time to reach the peak (v_0/g), not a valid root for this height.

1. Vertical position with upward positive:
   $y = v_0 t - \tfrac{1}{2}g t^2 = 30t - 4.905 t^2$
2. Set $y = 25$ and rearrange:
   $4.905 t^2 - 30 t + 25 = 0$
3. Quadratic formula:
   $t = \frac{30 \pm \sqrt{30^2 - 4(4.905)(25)}}{2(4.905)} = \frac{30 \pm \sqrt{409.5}}{9.81}$
4. The two roots are $t = 1.00\,\text{s}$ (up) and $t = 5.12\,\text{s}$ (down). Take the downward pass:
   $t = \frac{30 + 20.24}{9.81} = 5.12\,\text{s}$

FE Reference Handbook 10.6, p. 106

No time is given, so use the angular no-time equation $\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)$, the rotational twin of $v^2 = v_0^2 + 2a\,s$. The pulley stops, so $\omega = 0$. Solve for $\theta = \omega_0^2 / (2\alpha) = 1600 / 5 = 320\,\text{rad}$. The $16\,\text{rad}$ option is $\omega_0 / \alpha$, the time to stop, not the angle. The $640\,\text{rad}$ option drops the factor of 2 in the denominator ($1600/2.5 = 640$). The $160\,\text{rad}$ option is just half of the correct answer.

1. Angular no-time equation, stopping so $\omega = 0$:
   $\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)$
2. Insert values with $\alpha = -2.5\,\text{rad/s}^2$:
   $0 = 40^2 + 2(-2.5)\,\theta$
3. Solve for the angle:
   $\theta = \frac{1600}{5} = 320\,\text{rad}$

FE Reference Handbook 10.6, p. 110

A tangent axis is one radius away from the center, so $d = R = 0.10\,\text{m}$. Compute the centroidal term $\tfrac{2}{5}(6)(0.10^2) = 0.024$, then add the transfer term $md^2 = 6(0.10^2) = 0.060$. Total is $0.084\,\text{kg}{\cdot}\text{m}^2$. The 0.024 option keeps only the centroidal term. The 0.060 option keeps only the transfer term. The 0.216 option uses $d = 2R$ (diameter) instead of $d = R$ for the tangent distance.

1. Centroidal MOI of the sphere:
   $I_c = \tfrac{2}{5}mR^2 = \tfrac{2}{5}(6)(0.10^2) = 0.024\,\text{kg}{\cdot}\text{m}^2$
2. Tangent distance equals the radius:
   $d = R = 0.10\,\text{m}$
3. Parallel-axis theorem:
   $I = I_c + md^2 = 0.024 + 6(0.10^2) = 0.024 + 0.060 = 0.084\,\text{kg}{\cdot}\text{m}^2$

FE Reference Handbook 10.6, p. 112

Treat both blocks as one system. The net driving force is the weight difference $(m_1 - m_2)g$, and the total mass being accelerated is $m_1 + m_2$. So $a = (m_1 - m_2)g / (m_1 + m_2) = (2)(9.81)/12 = 1.64\,\text{m/s}^2$. The 2.81 m/s² distractor divides the weight difference by only the heavier mass (19.62/7). The 3.92 m/s² distractor divides by 5, the lighter mass (19.62/5). The 0.33 m/s² distractor comes from dividing the mass difference by the lighter mass (2/6) instead of using the correct Atwood formula.

1. Driving force from the weight difference:
   $F_{net} = (m_1 - m_2)g = (7 - 5)(9.81) = 19.62\,\text{N}$
2. Total mass in motion:
   $m_{total} = m_1 + m_2 = 7 + 5 = 12\,\text{kg}$
3. Apply Newton’s second law to the system:
   $a = \frac{(m_1 - m_2)g}{m_1 + m_2} = \frac{19.62}{12} = 1.64\,\text{m/s}^2$

FE Reference Handbook 10.6, p. 106

The scale reads the normal force $N$. With upward positive, $N - mg = ma$, so $N = m(g + a) = 70(9.81 + 2.0) = 827\,\text{N}$. Accelerating upward makes you feel heavier. The $687\,\text{N}$ choice is the static weight $mg$, ignoring the acceleration. The $547\,\text{N}$ choice subtracts the acceleration term (as if descending). The $140\,\text{N}$ choice is just $ma$, the extra force alone, not the total reading.

1. Free body of the person, upward positive:
   $N - mg = ma$
2. Solve for the normal force:
   $N = m(g + a) = 70(9.81 + 2.0)$
3. Evaluate:
   $N = 70(11.81) = 827\,\text{N}$

FE Reference Handbook 10.6, p. 106

Change in kinetic energy is $\tfrac{1}{2}m(v_2^2 - v_1^2)$. Square each speed separately first: $25^2 - 10^2 = 625 - 100 = 525$. Then $\tfrac{1}{2}(1200)(525) = 315{,}000\,\text{J}$. The $135\,\text{kJ}$ choice squares the velocity difference $(25-10)^2 = 225$ instead of the difference of squares, giving $\tfrac{1}{2}(1200)(225) = 135{,}000\,\text{J}$. The $162\,\text{kJ}$ choice comes from a partial-squaring slip. The $270\,\text{kJ}$ choice both drops the one-half and uses the squared difference: $(1200)(225) = 270{,}000\,\text{J}$.

1. Change in kinetic energy:
   $\Delta T = \tfrac{1}{2}m(v_2^2 - v_1^2)$
2. Difference of squares (not the square of the difference):
   $v_2^2 - v_1^2 = 25^2 - 10^2 = 625 - 100 = 525\,\text{m}^2/\text{s}^2$
3. Evaluate:
   $\Delta T = \tfrac{1}{2}(1{,}200)(525) = 315{,}000\,\text{J} = 315\,\text{kJ}$

FE Reference Handbook 10.6, p. 107

At constant velocity, kinetic energy does not change, so by the work-energy theorem the NET work is zero. The applied force does positive work, but friction does an equal amount of negative work, dissipating it as heat. So the applied work equals the friction loss. The choice claiming kinetic energy increases is wrong because constant velocity means constant KE. The choice claiming no work is done is wrong: the applied force does work (force times displacement), it is just cancelled by friction. The choice claiming potential energy increases is wrong because horizontal motion does not change gravitational PE.

1. Work-energy theorem:
   $U_{net} = \Delta T = T_2 - T_1$
2. Constant velocity means constant kinetic energy:
   $\Delta T = 0 \implies U_{net} = 0$
3. The positive applied work is exactly balanced by negative friction work:
   $U_{applied} = -U_{friction} = F_f \, d$

FE Reference Handbook 10.6, p. 107

Impulse equals the change in momentum, and direction matters. Taking the outgoing direction as positive, the incoming velocity is -40 and the outgoing is +50, so the velocity change is $50 - (-40) = 90\,\text{m/s}$. Impulse is $m\,\Delta v = 0.15(90) = 13.5\,\text{N}{\cdot}\text{s}$. The 1.5 N·s option uses the speed difference $50 - 40 = 10$, ignoring the reversal. The 6.0 N·s option uses the incoming speed only ($0.15 \times 40$). The 7.5 N·s option uses the outgoing speed only ($0.15 \times 50$). The reversal is the whole point.

1. Take the outgoing direction as positive. Incoming velocity is negative:
   $v_1 = -40\,\text{m/s}, \quad v_2 = +50\,\text{m/s}$
2. Velocity change:
   $\Delta v = v_2 - v_1 = 50 - (-40) = 90\,\text{m/s}$
3. Impulse equals change in momentum:
   $J = m\,\Delta v = 0.15(90) = 13.5\,\text{N}{\cdot}\text{s}$

FE Reference Handbook 10.6, p. 108

The system starts with zero total momentum, and with no friction it stays zero. So the skater’s backward momentum must equal the ball’s forward momentum: $m_s v_s = m_b v_b$, giving $v_s = (3)(8)/60 = 0.40\,\text{m/s}$. The 0.13 m/s choice divides the ball speed by the skater mass ($8/60 \approx 0.13$) instead of using momentum. The 8.0 m/s choice just reports the ball speed. The 2.67 m/s choice divides the ball speed by 3 (8/3) instead of conserving momentum.

1. Conservation of momentum from rest (total momentum = 0):
   $0 = m_b v_b - m_s v_s$
2. Solve for the skater’s recoil speed:
   $v_s = \frac{m_b v_b}{m_s} = \frac{3 \times 8}{60}$
3. Evaluate:
   $v_s = \frac{24}{60} = 0.40\,\text{m/s}$

FE Reference Handbook 10.6, p. 109

For a simple pendulum the natural circular frequency is $\omega_n = \sqrt{g/L} = \sqrt{9.81/0.80} = 3.50\,\text{rad/s}$. The question asks for frequency in Hz, so divide by $2\pi$: $f_n = 3.50/(2\pi) = 0.56\,\text{Hz}$. The $3.50\,\text{Hz}$ choice reports $\omega_n$ in rad/s without converting. The $1.79\,\text{Hz}$ choice uses $\sqrt{L/g}$ inverted then a slip. The $0.28\,\text{Hz}$ choice halves the correct value (divides $\omega_n$ by an extra factor).

1. Natural circular frequency of a simple pendulum:
   $\omega_n = \sqrt{\frac{g}{L}} = \sqrt{\frac{9.81}{0.80}} = \sqrt{12.26} = 3.50\,\text{rad/s}$
2. Convert to Hz:
   $f_n = \frac{\omega_n}{2\pi} = \frac{3.50}{2\pi} = 0.56\,\text{Hz}$

FE Reference Handbook 10.6, p. 113

Springs in series get softer: the equivalent stiffness is $k_{eq} = (1/k_1 + 1/k_2)^{-1} = (1/1200 + 1/800)^{-1} = 480\,\text{N/m}$. Then $\omega_n = \sqrt{k_{eq}/m} = \sqrt{480/5} = \sqrt{96} = 9.80\,\text{rad/s}$. The $20.0\,\text{rad/s}$ option wrongly adds the stiffnesses (that is the parallel rule, giving $2000\,\text{N/m}$, then $\sqrt{2000/5}=20.0$). The $12.6\,\text{rad/s}$ option uses only $k_2$ ($\sqrt{800/5} = \sqrt{160} \approx 12.6$). The $6.93\,\text{rad/s}$ option comes from a softer-stiffness slip ($\sqrt{240/5} = \sqrt{48} \approx 6.93$).

1. Equivalent stiffness for series springs:
   $k_{eq} = \left(\frac{1}{k_1} + \frac{1}{k_2}\right)^{-1} = \left(\frac{1}{1{,}200} + \frac{1}{800}\right)^{-1}$
2. Evaluate the series combination:
   $k_{eq} = \left(\frac{2}{2{,}400} + \frac{3}{2{,}400}\right)^{-1} = \frac{2{,}400}{5} = 480\,\text{N/m}$
3. Natural circular frequency:
   $\omega_n = \sqrt{\frac{k_{eq}}{m}} = \sqrt{\frac{480}{5}} = \sqrt{96} = 9.80\,\text{rad/s}$

FE Reference Handbook 10.6, p. 113