Work and Energy Work is the product of Force and displacement. The force needed to calculate Work is the force or component force in the direction of the distance covered! Work is the dot product (scalar value) of force and displacement: W = F s Work (and Energy) is measured in Joules (J)! Work does not have direction and so it does not have x, y, and z components!!
As a dot product, the amount of work done is found: W = F s cos ø where ø is the angle between F and s
A block of mass 11.7 kg is to be pushed 4.65 m up the surface of an incline so that it is raised to a height of 2.86 m in the process. Assuming a frictionless surface, how much work is done by P pushing the block at constant speed? P 4.65 m 2.86 m
mg P ø N∑F x = 0 = P - mgsinø ∑Fy = 0 = N - mgcosø W = Ps = mgsinø(s) = (11.7)(9.80)(2.86/4.65)(4.65) = 328 J OR: Work is done going against gravity: W = mgh= (11.7)(9.8)(2.86)= 328 J The same amount of work is done in either case! mgcosø mgsinø
A child pulls a 5.6 kg sled a distance of 12 m along a horizontal surface at constant speed. What work does the child do to the sled if µ k =.20 and the cord the child uses to pull the sled is at an angle of 45˚ with the horizontal? mg P ø N fkfk ∑F x = 0 = Pcosø - f k ∑F y = 0 = N + Psinø - mg W = Pcosø(s) f k = Pcosø f k = µ k N
N = mg - Psinø Pcosø = µ(mg - Psinø)= µmg - Pµsinø P = µmg cosø + µsinø = 13 N W = Pcosø(s) = (13)cos(45˚)(12)= 110 J Work done by friction? W = f k s = µ(mg – Psinø 1 )s(cosø 2 ) = J ø 2 is 180˚, the angle between the force and the displacement vectors! Neg. Work = Energy Lost
Work done by a variable force in one dimension First consider a simple case of a constant force applied in the constant x direction: 15 N applied for 10 m: F(N) s(m) W = Fs = (15)(10) = 150 J The same as the highlighted area under the graph 150 J! W
With a force that varies: F s xx F1F1 F2F2 WW The rectangular area gives a good approximation of the work! As x gets small ( 0) the area comes closer to the instantaneous work!
F s If we could divide the area under the curve into an infinite amount of skinny rectangles, the sum of the area of those rectangles would be the area under the curve the work done by F in terms of x! W = F 1 x + F 2 x + F 3 x …. x0x0 x
The number of intervals would go to infinity, so the exact result would be written: W = lim ∑ F n x x0x0 This is the process of integration, finding an infinite amount of sums from one value to the next. We write integrals using the following notation: W = ∫ F(x) dx xixi xfxf
Integrating is the opposite process of taking a derivative: x(t) = x 2 dx/dt = 2x∫ (2x)dx = x 2 Therefore, ∫ x n = x n+1 n+1 Section 7.3 to 7.5 in the text provides a good discussion! A simple case to examine would be a particle attached to a spring. The amount of force needed to stretch a spring is found by F = kx, where k is the spring constant.
How much work is done in stretching a spring with constant 125 N/m from 10.0 cm to 30.0 cm? F = kx W = ∫ F(dx) = kx 2 — kx =.5(125)(.300) 2 -.5(125)(.100) 2 = 5.00 J The spring would do J of work in restoring the object from 30 cm to 10 cm!
A spring hangs vertically in equilibrium when a mass of 6.40 kg is attached to it by hand and slowly lowered at constant speed until equilibrium is again reached. At this point, the spring has been stretched a total distance of.124 m. Calculate the work done on the block by A) gravity B)the spring and C) the hand. A) W g = mgs = (6.40)(9.80)(.124) = 7.78 J* *the force and displacement are in the same direction (down) and so work is positive!
B) W s = ∫ -ks = - ks 2 /2 k = F/s = mg/s = (6.40 kg)(9.80 m/s 2 ) (.124 m) = 506 N/m W s = -.5 (506)(.124) 2 = J * *negative because it is opposite of the displacement! The work done by the hand in lowering the the mass can be found in two ways:
A) If the block is lowered at constant speed, then: ∑F y = 0 = mg - (F s + F h ) = mg - ks - F h F h = mg - ks W h = - ∫ F h ds = - ∫ (mg - ks)ds = - mgs + ks 2 /2 W h = = J B) The total work done must be 0 because we are in equilibrium: W net = W s + W h + W g 0 = (-3.89 J) + W h + (7.78 J) W h = J 0 s
Work can be computed in terms of any force acting upon an object: If an object accelerates, then net Work is done: W = F∆x = ma∆x v 2 – v 0 2 = 2a∆x a∆x =.5(v 2 – v 0 2 ) W =.5m(v 2 – v 0 2 )
Kinetic Energy and the Work-Energy Theorem If an object is acted upon by a F net then it will accelerate! That net force will do net work on the object! The net work done will be equal to the change in kinetic energy of the object! W net = K f - K i =.5mv f 2 -.5mv i 2 =.5m(v f 2 - v i 2 ) No object can accelerate without work being done!
A body of mass 4.5 kg is dropped from a height of 10.5 m above the ground. Use the work- energy theorem to find the speed of the object as it hits the ground. W = F s = F w h = mgh W = ∆K = K f - K i =.5mv 2 -.5mv o 2 mgh =.5mv 2 v = √ 2gh = 14.3 m/s obviously, the same equation and result as if we had solved it as a simple freefall problem!
A block of mass 3.63 kg slides on a horizontal, frictionless table with a speed of 1.22 m/s. It is brought to rest by compressing a spring with force constant 135 N/m. How much was the spring compressed? ∆K = K f - K i =.5mv 2 –.5mv o 2 W = –.5kd 2 W = ∆K –.5kd 2 = –.5mv o 2 d = v o √ m/k =.200 m
A ball loses 15.0 % of its kinetic energy when it bounces back from a concrete walk. With what speed must you throw it vertically down from a height of 12.4 m in order to have it bounce back to the same height? A 125 g frisbee is thrown from a height of 1.06 m above the ground with a speed of 12.3 m/s. When it has reached a height of 2.32 m, its speed is 9.57 m/s. A) How much work was done on the frisbee by gravity? B) How much kinetic energy was lost due to air drag?
Power Power is the timed rate at which work is done: Average Power is total work over total time: P = W t Instantaneous Power: P = dW dt P = F v this would give instantaneous power at a given F and v, and constant power at constant F and v!