FM1-2 Work, energy and power — coverage pack

1 specification leaves · notes, questions, answers and worked methods

FM1-2.1 · Kinetic and potential energy, work and power. The work-energy principle. The principle of conservation of mechanical energy.

Kinetic energy is $\tfrac12mv^2$ , gravitational potential energy changes by $mg\Delta h$ , and work is $\int F\,dx$ when force varies with position.
The work-energy principle states that the work done by the resultant force equals the change in kinetic energy.
Mechanical energy is conserved when only conservative forces do work; resistance or driving forces must instead be included through their work.
Power is the rate of doing work: $P=dW/dt=Fv$ when the force and velocity are parallel. Do not use $P=Fv$ with an un-resolved force component.

1. A $1200\,\text{kg}$ car increases its speed from $10\,\text{m s}^{-1}$ to $14\,\text{m s}^{-1}$ . Calculate the net work done on the car.[2 marks]
Answer
- $57\,600\,\text{J}$
Method: Net work equals the gain in kinetic energy: $W=\tfrac12(1200)(14^2-10^2)=600(96)=57\,600\,\text{J}$ .

1. A $1200\,\text{kg}$ car climbs a road inclined at $5^\circ$ to the horizontal at a constant speed of $20\,\text{m s}^{-1}$ . The resistance is $400\,\text{N}$ . Taking $g=9.8\,\text{m s}^{-2}$ , find the engine power.[4 marks]
Answer
- $2.85\times10^4\,\text{W}$ , or $28.5\,\text{kW}$ (3 s.f.)
Method: Constant speed gives zero resultant force along the slope. The driving force is $F=1200(9.8)\sin5^\circ+400=1424.9\,\text{N}$ . Therefore $P=Fv=1424.9(20)=2.85\times10^4\,\text{W}$ .

1. A $2\,\text{kg}$ particle moves along a horizontal line. A constant driving force of $20\,\text{N}$ acts forwards while the resistance after displacement $x\,\text{m}$ is $(2+0.5x)\,\text{N}$ . Its speed at $x=0$ is $3\,\text{m s}^{-1}$ . Find its speed when $x=8$ .[5 marks]
Answer
- $\sqrt{137}\,\text{m s}^{-1}\approx11.7\,\text{m s}^{-1}$
Method: The driving work is $20(8)=160\,\text{J}$ . The work against resistance is $\int_0^8(2+0.5x)\,dx=[2x+0.25x^2]_0^8=32\,\text{J}$ . Hence the net work is $128\,\text{J}$ . By work-energy, $\tfrac12(2)v^2-\tfrac12(2)(3^2)=128$ , so $v^2=137$ and $v=\sqrt{137}\,\text{m s}^{-1}$ .