1. **Problem Statement:** Explain key physics terms and solve related calculations including work, kinetic energy, power, stopping time, distance under acceleration, work done lifting a load, and power output. 2. **Definitions:** i. Work: Work is defined as the product of force applied on an object and the displacement in the direction of the force, mathematically $$W = F imes d$$, measured in joules (J). ii. Kinetic Energy: The energy possessed by a body due to its motion, given by $$KE = \frac{1}{2}mv^2$$ where $m$ is mass and $v$ velocity. iii. Power: Power is the rate at which work is done or energy is transferred, mathematically $$P = \frac{W}{t}$$, measured in watts (W). 3. **Work-Energy Theorem:** States that the net work done on an object is equal to the change in its kinetic energy, $$W_{net} = \Delta KE$$. This signifies that work results in changes in the object's speed or motion energy. 4. **Calculations:** (b)(i) Calculate Kinetic Energy of car mass $m=1500$ kg, velocity $v=20$ m/s: $$KE = \frac{1}{2} \times 1500 \times 20^2 = 0.5 \times 1500 \times 400 = 300000\text{ J}$$ (c)(i) Truck velocity $v_i = 22.5$ m/s, deceleration $a = -2.27$ m/s$^2$; final velocity $v_f=0$. Time to stop $$t = \frac{v_f - v_i}{a} = \frac{0 - 22.5}{-2.27} = \frac{-22.5}{-2.27} \approx 9.91\text{ s}$$ (d)(i) Car initial velocity $u=30$ m/s, acceleration $a=-2$ m/s$^2$, final velocity $v=15$ m/s. Distance covered: Using $$v^2 = u^2 + 2as$$, $$15^2 = 30^2 + 2 (-2) s$$ $$225 = 900 - 4s$$ $$4s = 900 - 225 = 675$$ $$s = \frac{675}{4} = 168.75\text{ m}$$ (c)(ii) Crane lifting mass $m=500$ kg, height $h=15$ m, time $t=10$ s. (i) Work done against gravity: $$W = mgh = 500 \times 9.8 \times 15 = 73500\text{ J}$$ (ii) Average power output: $$P = \frac{W}{t} = \frac{73500}{10} = 7350\text{ W}$$