Forces, Acceleration & Energy

By Newton's second law, $a = \dfrac{\sum F}{m} = \dfrac{20}{4.0} = 5.0\text{ m/s}^2$.

$\sum F = ma = (2.0)(3.0) = 6.0\text{ N}$.

$KE = \tfrac{1}{2}mv^2 = \tfrac{1}{2}(2.0)(6.0)^2 = \tfrac{1}{2}(2.0)(36) = 36\text{ J}$. (Forgetting the $\tfrac{1}{2}$ gives the $72\text{ J}$ trap.)

Work is force times distance moved in the force's direction: $W = Fs = (15)(4.0) = 60\text{ J}$.

Power is the rate of doing work: $P = \dfrac{W}{t} = \dfrac{600}{5.0} = 120\text{ W}$.

By Newton's first law, an object keeps moving at constant velocity unless a net force acts. With no friction or gravity in deep space, no force is needed to maintain the motion, the engines are only needed to change it.

Find the net force first: $\sum F = 30 - 10 = 20\text{ N}$. Then $a = \dfrac{\sum F}{m} = \dfrac{20}{4.0} = 5.0\text{ m/s}^2$. (Using $20\text{ N}$ as the acceleration forgets to divide by the mass.)

On level ground $N = mg$, so $f = \mu N = \mu mg = (0.30)(10)(9.8) \approx 29\text{ N}$. (Forgetting $\mu$ gives the $98\text{ N}$ trap.)

First find the acceleration of the whole system: $a = \dfrac{F}{m_1 + m_2} = \dfrac{12}{3.0} = 4.0\text{ m/s}^2$. The only horizontal force on the $1.0\text{ kg}$ block is the contact push, so $F_{\text{contact}} = m_2 a = (1.0)(4.0) = 4.0\text{ N}$. (The full $12\text{ N}$ does not transmit, only enough to accelerate the second block.)

$\mu = \dfrac{f}{N} = \dfrac{f}{mg} = \dfrac{49}{(20)(9.8)} = \dfrac{49}{196} = 0.25$. The coefficient of friction has no units.

On a frictionless slope the acceleration is $a = g\sin\theta = (9.8)(0.5) = 4.9\text{ m/s}^2$. The mass cancels, so every object slides at the same rate.

Only the horizontal part of the force does work over the horizontal displacement: $W = Fs\cos\theta = (50)(10)(0.87) \approx 435\text{ J}$. (Ignoring the angle gives the $500\text{ J}$ trap.)

By the work-energy theorem the work equals the gain in kinetic energy: $W = \Delta KE = \tfrac{1}{2}mv^2 - 0 = \tfrac{1}{2}(1000)(20)^2 = 2.0\times10^5\text{ J}$. (Dropping the $\tfrac{1}{2}$ gives $4.0\times10^5$.)

For a force moving at a steady speed, $P = Fv = (500)(30) = 15{,}000\text{ W}$ ($15\text{ kW}$).

Kinetic energy depends on the square of the speed ($KE = \tfrac{1}{2}mv^2$), so doubling $v$ multiplies $KE$ by $2^2 = 4$. (This is why stopping distances grow so quickly with speed.)

The net upward force gives the acceleration: $N - mg = ma$, so $N = m(g + a) = 60(9.8 + 2.0) = 60(11.8) = 708\text{ N}$. Accelerating upward, the person feels heavier than their true weight of $588\text{ N}$.

Down the slope: $ma = mg\sin\theta - \mu mg\cos\theta$, so $a = g(\sin\theta - \mu\cos\theta) = 9.8(0.5 - 0.2 \times 0.87) = 9.8(0.326) \approx 3.2\text{ m/s}^2$. (Ignoring friction gives $4.9\text{ m/s}^2$.)

The hanging weight drives the whole $5.0\text{ kg}$ system: $a = \dfrac{m_2 g}{m_1 + m_2} = \dfrac{(2.0)(9.8)}{3.0 + 2.0} = \dfrac{19.6}{5.0} \approx 3.9\text{ m/s}^2$. (Both masses accelerate together, so the total mass must be used.)

The braking work removes all the kinetic energy: $Fd = \tfrac{1}{2}mv^2$, so $d = \dfrac{\tfrac{1}{2}mv^2}{F} = \dfrac{\tfrac{1}{2}(1200)(20)^2}{6000} = \dfrac{240{,}000}{6000} = 40\text{ m}$.

The work done against gravity is $W = mgh = (500)(9.8)(20) = 98{,}000\text{ J}$, delivered in $10\text{ s}$: $P = \dfrac{W}{t} = \dfrac{98{,}000}{10} = 9800\text{ W}$. (Forgetting to divide by the time gives $98{,}000$.)

The vertical drop is $h = L\sin\theta = 5.0 \times 0.5 = 2.5\text{ m}$. Energy conservation gives $\tfrac{1}{2}mv^2 = mgh$, so $v = \sqrt{2gh} = \sqrt{2(9.8)(2.5)} = \sqrt{49} = 7.0\text{ m/s}$. (Using the slope length $5\text{ m}$ as the height gives the $9.9\text{ m/s}$ trap.)

Taking down as the acceleration direction: $mg - N = ma$, so $N = m(g - a) = 70(9.8 - 3.0) = 70(6.8) = 476\text{ N}$. Accelerating downward, the person feels lighter than their true weight of $686\text{ N}$.

The net work equals the change in kinetic energy: $Fd = \tfrac{1}{2}m(v^2 - u^2) = \tfrac{1}{2}(1000)(30^2 - 10^2) = \tfrac{1}{2}(1000)(800) = 4.0\times10^5\text{ J}$. Then $F = \dfrac{4.0\times10^5}{100} = 4000\text{ N}$.

At constant speed the driving force equals the drag ($800\text{ N}$), so $P = Fv = (800)(25) = 20{,}000\text{ W}$ ($20\text{ kW}$). All of this power ends up overcoming drag.

The energy lost is the drop in gravitational potential energy between the two heights: $\Delta E = mg(h_1 - h_2) = (0.50)(9.8)(8.0 - 5.0) = (0.50)(9.8)(3.0) = 14.7\text{ J}$. (The full $mgh_1 = 39.2\text{ J}$ is the energy it started with, not the energy lost.)