Kinetics of Particles: Work Energy Principles

(a) Principle of work and energy

Statement: the work done by all forces acting on a particle as it moves from position 1 to position 2 equals the change in its kinetic energy.

$$U_{1\to2} = T_2 - T_1 = \tfrac12 m v_2^2 - \tfrac12 m v_1^2$$

Proof: Let the resultant force $\mathbf{F}$ act on a particle of mass $m$ moving along a path with tangential acceleration $a_t$. Newton's second law along the tangent gives $F_t = m a_t = m\,\dfrac{dv}{dt}$. Using $a_t = v\,\dfrac{dv}{ds}$:

$$F_t\,ds = m v\,dv$$

Integrating from position 1 ($s_1$, $v_1$) to position 2 ($s_2$, $v_2$):

$$\int_{s_1}^{s_2} F_t\,ds = m\int_{v_1}^{v_2} v\,dv = \tfrac12 m v_2^2 - \tfrac12 m v_1^2$$

The left side is the work $U_{1\to2}$ of the resultant force, so $T_1 + U_{1\to2} = T_2$. Normal forces do no work as they are perpendicular to the displacement.

(b) Power and efficiency

Power is the rate of doing work:

$$P = \frac{dU}{dt} = \mathbf{F}\cdot\mathbf{v} = Fv\cos\alpha$$

where $\alpha$ is the angle between the force and the velocity. Unit: watt (W = J/s); 1 hp = 746 W. For rotation, $P = T\omega$ (torque times angular velocity).

Mechanical efficiency is the ratio of useful output power (or work) to input power (or work):

$$\eta = \frac{\text{output power}}{\text{input power}} = \frac{P_{out}}{P_{in}} < 1$$

The difference between input and output is lost mainly to friction and appears as heat.

Given: $m = 10$ kg, $\theta = 30^\circ$, $\mu = 0.20$, $d = 4$ m, $k = 2000$ N/m.

   \
    \  block
     \[#]
      \ \  4 m
       \ \_____
        \  /\/\/\ spring
         \/_______

Forces along the incline

• Component of weight down the slope: $mg\sin30^\circ = 10 \times 9.81 \times 0.5 = 49.05$ N
• Normal reaction: $N = mg\cos30^\circ = 84.96$ N
• Friction (up the slope): $\mu N = 0.20 \times 84.96 = 16.99$ N
• Net driving force: $49.05 - 16.99 = 32.06$ N

Speed just before touching the spring

Work-energy over 4 m: $0 + F_{net}\, d = \tfrac12 m v^2$

$$v = \sqrt{\frac{2 \times 32.06 \times 4}{10}} = 5.06\ \text{m/s}$$

Maximum compression x

The block moves $(4 + x)$ along the incline from rest to rest (momentarily stopped at maximum compression). The spring does negative work $-\tfrac12 kx^2$.

$$F_{net}(4 + x) - \tfrac12 k x^2 = 0$$ $$1000x^2 - 32.06x - 128.23 = 0$$ $$x = \frac{32.06 + \sqrt{32.06^2 + 4(1000)(128.23)}}{2000} = 0.374\ \text{m}$$

(The negative root is rejected.)

Answer: Maximum compression = 374 mm (0.374 m); speed before touching the spring = 5.06 m/s.

(a) Conservative force and potential energy

A conservative force is a force whose work between two points depends only on the end positions and not on the path followed (for example gravity and spring force). Work done around any closed path is zero. Friction is non-conservative.

Potential energy $V$ is the energy a body has because of its position or configuration. The work of a conservative force equals the decrease in potential energy: $U_{1\to2} = V_1 - V_2$.

• Gravitational: $V_g = mgh$ (h measured from a chosen datum)
• Elastic (spring of stiffness $k$, extension $x$): $V_e = \tfrac12 kx^2$

Conservation of mechanical energy: if only conservative forces do work, the total mechanical energy remains constant:

$$T_1 + V_1 = T_2 + V_2$$

(b) Pendulum bob

Take the lowest point as datum. Tension does no work (it is perpendicular to the path), so energy is conserved.

   O ____________ (start, v = 0)
   |  \
   |   \ L = 1.2 m
   |    \
   v     o   <- lowest point

$$mgL = \tfrac12 m v^2 \;\Rightarrow\; v = \sqrt{2gL} = \sqrt{2 \times 9.81 \times 1.2} = 4.85\ \text{m/s}$$

At the lowest point the net upward force provides the centripetal force:

$$T - mg = \frac{mv^2}{L}$$ $$T = 0.5 \times 9.81 + \frac{0.5 \times (4.852)^2}{1.2} = 4.905 + 9.810 = 14.71\ \text{N}$$

This equals $3mg$, a known result for release from the horizontal.

Answer: v = 4.85 m/s; T = 14.71 N (= 3mg).

Given: $m = 1500$ kg, gradient 1 in 20 (take $\sin\theta = 1/20 = 0.05$), $v = 54$ km/h $= 54/3.6 = 15$ m/s, resistance $R = 400$ N, $\eta = 0.80$.

At constant speed the driving force along the road balances the resistance and the weight component down the slope.

$$\begin{aligned} mg\sin\theta &= 1500 \times 9.81 \times 0.05 = 735.75\ \text{N} \\ F &= R + mg\sin\theta = 400 + 735.75 = 1135.75\ \text{N} \end{aligned}$$

Power at the driving wheels

$$P_{wheels} = Fv = 1135.75 \times 15 = 17036\ \text{W} = 17.04\ \text{kW}$$

Engine power

Efficiency $= P_{wheels}/P_{engine}$:

$$P_{engine} = \frac{17.04}{0.80} = 21.30\ \text{kW} \;(\approx 28.5\ \text{hp})$$

Answer: Power at wheels = 17.04 kW; engine power = 21.30 kW.