Class notes Physics

‭Work Power and Efficiency‬

‭ .‬‭A person pushes a 10.0 kg mass at constant speed‬‭across μ = 0.150, 8.50 m long floor‬
1
‭at constant speed.‬
‭a)‬ ‭How much work does the person do?‬
‭b)‬ ‭How much work does friction do?‬
‭Ans:‬
‭For part A, since the person is pushing the mass with a‬‭constant velocity‬‭, which means‬
‭no acceleration, forces are balanced, and applied force is equal to friction force.‬
‭Givens:‬
→ →
‭𝐹‬‭𝑎𝑝‬ = ‭𝐹‬‭𝑓𝑟‬
→ →
‭𝐹‬‭𝑎𝑝‬ = ‭μ‬‭𝐹‭𝑁
‬‬
→ →
‭𝐹‬‭𝑎𝑝‬ = ‭μ‬‭𝑚‭𝑔
‬‬
→ ‭𝑁‬
‭𝐹‬‭𝑎𝑝‬ = ‭0‬. ‭150‬(‭10‬. ‭0‬‭𝑘𝑔‬)(‭9‬. ‭8‬ ‭𝑘𝑔‬ )
→
‭𝐹‬‭𝑎𝑝‬ = ‭14‬. ‭7‬‭𝑁‬
→→
‭Now, we have the applied force. To calculate the work, use the equation‬‭𝑊‬ = ‭𝐹‭𝑑
‬ ‬‭.‬
→→
‭ ‬ = ‭𝐹‭𝑑
𝑊 ‬‬
‭𝑊‬ = ‭14‬. ‭7‬‭𝑁‬(‭8‬. ‭50‬‭𝑚‬)
‭𝑊‬ = ‭125‬‭‬‭𝑁𝑚‬ ‭(‭s‬ ince‬‭𝑊‬ = ‭Δ‬‭𝐸‬, ‭which means the unit‬‭𝑁𝑚‬‭is equal to‬‭𝐽‬‭)‬
‭𝑊‬ = ‭125‬‭‬‭𝐽‬
‭For part B, since the applied force is equal to friction force because of constant velocity,‬
‭the work done by friction is also equal to‬‭125‬‭‬‭𝐽‬‭but negative, so the answer is‬− ‭125‬‭𝐽‬‭.‬

‭ .‬‭Find the power required for a crane to lift a‬‭400‬. ‭0‬‭𝑘𝑔‬‭mass at constant speed‬‭100‬. ‭0‬‭‬
2
‭meters in‬‭30‬. ‭0‬‭‬‭seconds. If the crane consumes‬‭35‬‭𝑘𝑊‬‭of electrical power, find its‬
‭efficiency.‬
‭Ans:‬
‭ ‬
𝑤
‭To find the power, use the equation‬‭𝑃‬ = ‭𝑡‬
‭. Since the crane lifts the mass at‬‭constant‬
s‭ peed‬‭, which means there is no acceleration. The applied‬‭force is equal to the gravity‬
‭force.‬
‭Givens:‬
‭ ‬
𝑤
‭𝑃‬ = ‭𝑡‬