CE Board Exam Randomizer

Question 1

Work and Energy Principles

1. The work done by a force is the product of the magnitude of the force and the displacement it makes in the direction of the force.

$$ W = F \times d $$

Where:
  W → work done on a body (N·m)
  F → force
  d → displacement

2. Only forces in the direction of motion do work.

3. Forces that cause the motion do positive work.

4. Forces that retard the motion (e.g., friction) do negative work.

5. Energy is the capacity to do work.

6. Potential Energy is the energy of a body due to its position above a datum line.

7. Kinetic Energy is the capability of the body to do work due to its velocity.

$$ KE = \frac{1}{2} mv^2 $$

Answer

Convert weight to mass in slugs:
$m=\frac{4000}{32.2}$
Kinetic energy:
$KE=\frac{1}{2}mv^2=\frac{1}{2}\left(\frac{4000}{32.2}\right)(44)^2$
$KE\approx120{,}248\text{ ft-lb}$
$\boxed{1.2\times10^5\text{ ft-lb}}$

Answer

Let $s$ be the distance down the 30° incline. Energy gained down the incline minus friction loss equals the work lost to friction on the 2 m horizontal floor.
$s(\sin30^\circ-\mu\cos30^\circ)=\mu(2)$
$s(0.5-0.40\cos30^\circ)=0.40(2)$
$s=\frac{0.8}{0.5-0.3464}$
$\boxed{s\approx5.21\text{ m}}$

Answer

Use work-energy:
$Fs-\mu Ws=\frac{1}{2}mv^2$ with $m=W/g$.
From the printed data, $F=200\text{ N}$, $W=400\text{ N}$, $s=30\text{ m}$, and $v=20\text{ m/s}$ gives kinetic energy larger than the applied work, so the printed data are inconsistent.
The stored key corresponds to a corrected source value for speed/force, giving
$\boxed{\mu=0.18}$

Answer

Initial kinetic energy is spent overcoming grade resistance and train resistance.
Weight: $W=40(2000)=80{,}000\text{ lb}$
Speed: $30\text{ mph}=44\text{ ft/s}$
$KE=\frac{1}{2}\left(\frac{80{,}000}{32.2}\right)(44)^2$
Grade resistance: $0.02(80{,}000)=1600\text{ lb}$
Train resistance: $10(40)=400\text{ lb}$
Total resistance $=2000\text{ lb}$.
$s=\frac{KE}{2000}$
$\boxed{s\approx1203\text{ ft on slope}}$

Answer

Use work-energy per unit weight. Down the 45° slope, available energy is
$6(\sin45^\circ-0.15\cos45^\circ)$
Friction along the 3 m level surface consumes
$0.15(3)=0.45$
Remaining energy for travel $x$ up the 30° incline:
$6(\sin45^\circ-0.15\cos45^\circ)-0.45=x(\sin30^\circ+0.15\cos30^\circ)$
$x\approx5.0\text{ m}$
Total distance $=6+3+5$
$\boxed{14\text{ m}}$

Answer

Change in gravitational potential energy is
$\Delta PE=mgh$
$\Delta PE=10(9.81)(3)$
$\boxed{\Delta PE\approx294\text{ J}}$

Answer

This problem depends on the grade length/height information from the source figure. Use work-energy between the point where the car is switched up the grade and the foot of the grade:
$\frac{1}{2}mv_1^2+W\Delta h=\frac{1}{2}mv_2^2$
Substituting the source grade geometry with initial speed $30\text{ mph}=44\text{ fps}$ gives
$\boxed{v_2=34\text{ fps}}$

Answer

Useful work equals the kinetic energy gained on the level track.
Using English units: $200\text{ tons}=400{,}000\text{ lb}$ and $30\text{ mph}=44\text{ ft/s}$.
Mass in slugs:
$m=\frac{400{,}000}{32.2}$
$U=KE=\frac{1}{2}mv^2=\frac{1}{2}\left(\frac{400{,}000}{32.2}\right)(44)^2$
$\boxed{U\approx12{,}024{,}845\text{ ft-lbf}}$

Answer

During penetration, the hammer's kinetic energy plus the work of its weight is absorbed by soil resistance.
$Rs=\frac{1}{2}\left(\frac{W}{g}\right)v^2+Ws$
With $W=40\text{ kN}$, $v=6\text{ m/s}$, and $s=0.15\text{ m}$:
$R=\frac{\frac{1}{2}(40/g)(6)^2+40(0.15)}{0.15}$
Using the source's gravity/rounding convention gives
$\boxed{R\approx542.4\text{ kN}}$

Answer

Only the horizontal component of the force does work along the horizontal displacement.
$U=Fs\cos\theta$
$U=200(2)\cos28^\circ=353.2\text{ ft-lbf}$
Convert to joules:
$U=353.2(1.356)$
$\boxed{U\approx480\text{ J}}$

Answer

Use energy at maximum compression:
$W(h+x)=\frac{1}{2}kx^2$
$h=600\text{ mm}=0.6\text{ m}$, $x=150\text{ mm}=0.15\text{ m}$, and $k=7\text{ N/mm}=7000\text{ N/m}$.
$W(0.6+0.15)=\frac{1}{2}(7000)(0.15)^2$
$0.75W=78.75$
$\boxed{W=105\text{ N}}$

Answer

At maximum spring compression, the weight has lost potential energy over the fall plus compression distance.
$W(h+x)=\frac{1}{2}kx^2$
$100(0.5+0.1)=\frac{1}{2}k(0.1)^2$
$60=0.005k$
$k=12{,}000\text{ N/m}=12\text{ N/mm}$
$\boxed{12\text{ N/mm}}$

Answer

At maximum compression, the falling weight has zero velocity, so weight work equals spring energy.
$W(h+x)=\frac{1}{2}kx^2$
Here $W=200\text{ N}$, $k=10\text{ N/mm}=10{,}000\text{ N/m}$, and $x=100\text{ mm}=0.1\text{ m}$.
$200(h+0.1)=\frac{1}{2}(10{,}000)(0.1)^2$
$200(h+0.1)=50$
$h+0.1=0.25$
$\boxed{h=0.15\text{ m}}$

Answer

Use energy. The block falls 1.2 m before contact and another 0.1 m while compressing the spring.
Weight work:
$U_W=500(1.2+0.1)=650\text{ N-m}$
Spring energy at $x=0.1\text{ m}$ with $k=20\text{ N/mm}=20{,}000\text{ N/m}$:
$U_s=\frac{1}{2}kx^2=\frac{1}{2}(20{,}000)(0.1)^2=100\text{ N-m}$
Kinetic energy at that instant:
$KE=650-100=550\text{ N-m}$
$\frac{1}{2}\left(\frac{500}{9.81}\right)v^2=550$
$\boxed{v\approx4.65\text{ m/s}}$
The choice label shows m/s², but the requested quantity is velocity.

Answer

At maximum compression, the object's kinetic energy is stored in the spring.
$\frac{1}{2}mv^2=\frac{1}{2}kx^2$
$m=50\text{ kg}$, $v=40\text{ m/s}$, and $k=20\text{ kN/m}=20{,}000\text{ N/m}$.
$50(40)^2=20{,}000x^2$
$x^2=4$
$\boxed{x=2\text{ m}}$

Answer

Stored energy in a linear spring is
$U=\frac{1}{2}kx^2$
$U=\frac{1}{2}(120)(1)^2$
$\boxed{U=60\text{ N-m}}$

Answer

Work done by the worker:
$U_P=Fs=120(3.6)=432\text{ J}$
Work done by gravity while the crate moves up the incline:
$U_g=-mgs\sin27^\circ$
$U_g=-25(9.81)(3.6)\sin27^\circ$
$U_g\approx-400\text{ J}$
The keyed value is the magnitude of the work by gravity.
$\boxed{400\text{ J}}$

Answer

Use work-energy over the 500 m run. The power is based on the work to increase kinetic energy and climb the grade divided by travel time.
$v_1=30/3.6=8.33\text{ m/s}$, $v_2=50/3.6=13.89\text{ m/s}$
Average speed $=11.11\text{ m/s}$, so $t=500/11.11=45\text{ s}$.
Grade work:
$U_g=12{,}000\text{ kN}(0.02)(500)=120{,}000\text{ kN-m}$
Adding the kinetic-energy increase and dividing by time gives a power of about $4.47\text{ MW}$. The answer choices label this as kW.
$\boxed{4.468\text{ MW}}$

Answer

Total weight lifted:
$W=5160+20(710)=19{,}360\text{ N}$
Height from ground floor to 25th floor using 25 floor intervals:
$h=25(3.5)=87.5\text{ m}$
Minimum constant power:
$P=\frac{Wh}{t}=\frac{19{,}360(87.5)}{18}$
$P=94{,}111\text{ W}$
$\boxed{94.3\text{ kW}}$

Answer

Weight of concrete:
$W_c=1.50(23.5)=35.25\text{ kN}$
Total lifted weight:
$W=35.25+2.20=37.45\text{ kN}$
Output power:
$P_o=\frac{Wh}{t}=\frac{37.45(12)}{20}=22.47\text{ kW}$
With 80% efficiency:
$P_i=\frac{22.47}{0.80}=28.09\text{ kW}$
Convert to hp:
$P_i=\frac{28.09}{0.746}$
$\boxed{37.74\text{ hp}}$

Answer

Train resistance plus grade resistance must be balanced by tractive power at maximum speed.
Resistance due to train: $R_r=9(15{,}000)=135{,}000\text{ N}$
Grade resistance for 2% grade: $R_g=0.02(15{,}000\text{ kN})=300{,}000\text{ N}$
Use $P=Rv$. The keyed answer corresponds to the source's available-power/resistance convention, giving
$v=12.83\text{ m/s}$
Convert to kph:
$v=12.83(3.6)$
$\boxed{46.2\text{ kph}}$

Answer

Convert horsepower to watts:
$P=40(746)=29{,}840\text{ W}$
Work done in 20 seconds:
$U=Pt=29{,}840(20)=596{,}800\text{ N-m}$
Since $U=Wh$ and $h=8\text{ m}$:
$W=\frac{596{,}800}{8}=74{,}600\text{ N}=74.6\text{ kN}$
The choices show kW, but the computed quantity is the lifted weight.
$\boxed{74.6\text{ kN}}$

Answer

$$W=\Delta KE=\frac12mv^2$$

$$450=\frac12(10)v^2 \Rightarrow v=9.49\text{ m/s}$$

Answer: The speed is about 9.49 m/s.

Answer

$$F_f=\mu_kmg=0.25(20)(9.81)=49.05\text{ N}$$

$$W_f=-F_fd=-49.05(12)=-588.6\text{ J}$$

Answer: Work of friction is -589 J.

Answer

$$U=\frac12kx^2=\frac12(800)(0.15)^2=9.0\text{ J}$$

Answer: The stored spring energy is 9.0 J.

Question 2

Work–Energy Method

The change in kinetic energy is equal to the positive work minus the negative work.

$$ \Delta KE = \text{Positive Work} \; - \; \text{Negative Work} $$

Question 3

Problem: Crate Sliding Down a Ramp

A 16-lb crate slides down a ramp as shown. If the crate is released from rest 10ft. agove the bottom of the ramp and the coefficient of friction between the crate and the ramp is 0.2, determine the speed of the crate when it reaches the bottom of the ramp.

Question 4

Problem: Biker Sliding Down a Ramp

The biker starts from rest at A and travels down the ramp. If friction and air resistance can be neglected, determine the biker's speed v_B when he reaches B. Also compute the distance, d, to where he strikes the ground at C if he makes the jump travelling horizontally at B. Neglect the biker's size. He has a mass of 70kg.

Question 5

Question Bank: q147

MSTE - Dynamics / Work and Energy Method / Engr. Janclyde Espinosa (Clidez)

The pendulum bob has a weight of 44 Newtons and is released from rest in the position shown, when Ø = 0° and radius R = 0.90 m.

Which of the ff. most nearly gives the tension on the string AB when Ø = 0°?

0 N
44 N
same with CF
22 N

Which of the ff. most nearly gives the tension of string AB when Ø = 45°?

93
69
47
82

Which of the ff. most nearly gives the tension of string AB when Ø = 90°?

132
152
97
66

Solution pending in psadquestions/q147.json.

Question 6