Answer to Question #166795 in Quantum Mechanics for Kipkirui Emmanuel

Question

The speed of a small truck increases from 36km/h to 72km/h During a three kilometer trip along a horizontal road.the mass of the truck is three thousand kilograms and the friction coefficient is 0.001.

1.) determine the work done and the average power developed by the trucks engine.

2.)same question if the truck travels uphill at an angle 10⁰

Answer 1

1) (a) Let's first find the work done in moving the truck by 3 km:

$$W=\Delta KE=KE_f-KE_i=\dfrac{1}{2}m(v_f^2-v_i^2),$$ $$W=\dfrac{1}{2}\cdot3000\ kg\cdot((20\ \dfrac{m}{s})^2-(10\ \dfrac{m}{s})^2)=450000\ J.$$

Then, let's calculate the work done against the force of friction:

$$W_{fr}=\mu mgdcos\theta,$$ $$W_{fr}=0.001\cdot3000\ kg\cdot9.8\ \dfrac{m}{s^2}\cdot3000\ m\cdot cos180^{\circ}=-88200\ J.$$

Finally, we can find the net work done by the trucks engine:

$$W_{net}=W-W_{fr}=450000\ J-88200\ J=361800\ J.$$

(b) Let's find the time that the truck takes to travel the distance of 3 kilometers:

$$d=\dfrac{1}{2}(v_0+v)t,$$ $$t=\dfrac{2d}{(v_0+v)},$$ $$t=\dfrac{2\cdot3000\ m}{(10\ \dfrac{m}{s}+20\ \dfrac{m}{s})}=200\ s.$$

Finally, we can find the average power developed by the trucks engine:

$$P_{avg}=\dfrac{W}{t}=\dfrac{361800\ J}{200\ s}=1809\ W.$$

2) (a) Let's calculate the work done against the force of friction in case of the truck travels uphill at an angle 10⁰:

$$W_{fr}=\mu mg cos\alpha d cos\theta,$$ $$W_{fr}=0.001\cdot3000\ kg\cdot9.8\ \dfrac{m}{s^2}\cdot cos10^{\circ} \cdot3000\ m\cdot cos180^{\circ}=-86860\ J.$$

Finally, we can find the net work done by the trucks engine:

$$W_{net}=W-W_{fr}=450000\ J-86860\ J=363140\ J.$$

(b) Finally, we can find the average power developed by the trucks engine:

$$P_{avg}=\dfrac{W}{t}=\dfrac{363140\ J}{200\ s}=1816\ W.$$