Take the W gravity for example… The equation is W gravity = mg(h-h 0 ) So if an object falls 50 meters, the work exerted on the object by gravity won’t change whether the object fall straight down or falls on an angle. All that matters is the initial and final heights.
Effectively, you can calculate the work exerted on an object by non-conservative forces by… finding the difference between the initial and final energies due to conservative forces acting on an object. The difference is the work due to non-conservative forces that acts on the object
So… Wnc = Δ KE + Δ PE + Δ PE s or W nc = ( ½mv² - ½ mv 0 ²) + (mgh – mgh 0 ) + (½ kx² - ½kx 0 ²) ( Δ means change)
An archer pulls a bowstring back a distance of 0.470m and then releases the arrow. The bow and string act like a spring whose spring constant is 425 N/m. What is the elastic potential energy of the system?
A motorcyclist is trying to leap across a canyon by driving horizontally off the cliff at a speed of 38.0 m/s. Ignoring air resistance, find the speed at which the motorcycle strikes the ground on the other side. The motorcycle starts at a height of 70.0m and will end at a height of 35.0m.
½ mv ² + mgh = ½mv 0 ² + mgh 0 (mass cancels out) ½v ² + (9.8m/s²)(35.0m) = ½(38.0m/s) ² + (9.8m/s²)(70.0m) V = 46.2 m/s
A 0.20kg rocket is launched from rest. It takes a roundabout route until it reaches a height of 29m above its starting point. In the process, 425J of work is done on the rocket by non-conservative forces (the burning propellant). Ignoring air resistance and the mass lost due to the burning of the fuel, find the speed of the rocket when it is 29 m above its starting point.
A car starts from rest and accelerates in a positive direction. The car has a mass of 1.10 x 10 ³kg and accelerates at +4.60m/s² for 5.00s. Determine the average power generated by the force that moves the car.
V = v 0 + at v = 0m/s + (4.6m/s)(5.00s) v = 23 m/s P = Fv P = mav P = (1.10 x 10 ³kg)(4.6m/s²)(23m/s) P = 1.16 x 10^5 W