4th Year Physics Revision Sheet

**4^rd^ Year Work and Power Revision Sheet** Energy is always conserved, but can be transferred from one form to another. It is measured in joules. Forces can transfer energy. The energy transferred in this way is called the **work done**. **Work Done \[J\] = Force\[N\] x Distance moved in the di...

**4^rd^ Year Work and Power Revision Sheet** Energy is always conserved, but can be transferred from one form to another. It is measured in joules. Forces can transfer energy. The energy transferred in this way is called the **work done**. **Work Done \[J\] = Force\[N\] x Distance moved in the direction of the force \[m\]** **W = F** x **d** e.g. If a fork lift truck does 1000J of work on a crate by lifting it, then the crate has gained 1000J of GPE. The work done **ON** an object is the energy transferred to it. (e.g. the fork lift did 1000J of work **on** the crate) The work done **BY** an object is the energy transferred from it. (e.g. 2000J of work done **by** the fork lift, converting chemical energy to GPE -- it is only 50% efficient) **Gravitational Potential Energy 'GPE'** The change in GPE of an object changing height can be calculated using: **Change in GPE \[J\] = mass\[kg\] x gravitational field strength\[N/kg\] x change in height \[m\]** **∆GPE = m** x **g** x **∆h** **Kinetic Energy** The total kinetic energy an object has (due to its motion) can be calculated using: Kinetic Energy \[J\] = ½ x mass \[kg\] x (speed \[m/s\])^2^ **KE** = **½** x **m** x **v^2^** e.g. **1)** A go kart of mass 40kg rolls from the top of a hill of height 5m. How fast is it travelling at the bottom of the hill? As the cart descends it loses GPE. ∆GPE = mg∆h = 40 x 10 x 5 = 2000J If there is no friction, then this will all become KE at the bottom. KE = ½ mv^2^ v = [\$\\sqrt{\\frac{2 \\times KE}{m}} = \\sqrt{\\frac{2\\ \\times \\ 2000}{40}}\$]{.math.inline} = **10m/s** **2)** If, when there is friction, the cart is only travelling at 7m/s, then how much work has been done by friction? The actual KE at the bottom = ½ mv^2^ = ½ x 40 x 7^2^ = 980J 2000J -- 980J = **1020J** of work done by friction. **Power** (measured in watts 'W', 1 watt = 1J/s.) Power measures the rate at which energy is transferred from one form to another. Two cars may each transfer 450kJ of chemical energy into KE when they accelerate. However if one manages this in only 3s, while the other takes 11s, then it is a more powerful car and it converts 150kJ/s = 150kW. **Power \[W\] =** [\$\\frac{\\mathbf{Work\\ done\\ \\lbrack J\\rbrack}}{\\mathbf{Time\\ taken\\lbrack s\\rbrack}}\\mathbf{= \\ }\\frac{\\mathbf{\\text{Energy\\ transferred}}}{\\mathbf{\\text{Time\\ taken}}}\$]{.math.inline} [\$\\mathbf{P = \\ }\\frac{\\mathbf{W}}{\\mathbf{t}}\$]{.math.inline}