Come in and turn your lab notebooks into the bin next to file cabinet. Work and Energy Come in and turn your lab notebooks into the bin next to file cabinet.

Energy Objects (or systems) can have energy and transfer or transform it. When Energy is transferred, we call it “doing work”. Since energy is conserved, we state that: W = ΔE

Work Work = F d cos (θ) F d F d Work is measured in: Remember: Work energy required to make something move. Angle b/t F and d Force (N) Work = F d cos (θ) Work Energy (J) Distance (m) Use only the magnitude of F and d in the equation. The angle will determine if work is positive or negative. Work is measured in: Newton meters (N *m) Joules (J) F d F d WORK DONE NO WORK

A shopper pushes a shopping cart on a rough surface with a force of 8 A shopper pushes a shopping cart on a rough surface with a force of 8.9 N at an angle of 60° to the left of the negative y- axis. While the cart moves a horizontal distance of 10.0 m, what is the work done by the shopper on the shopping cart? Fa = 8.9 N at an angle of 60° to the left of the negative y- axis Y-axis 8.9 N Θ=60° 10 meters

A 40-N force pulls a 4-kg block a horizontal distance of 8 m A 40-N force pulls a 4-kg block a horizontal distance of 8 m. The rope makes an angle of 350 with the floor and μk = 0.2. What is the work done by each force acting on the block AND what is the NET Work (total work)? x Fa q = 40 N = 8m = 35° 4kg μk= 0.2

Find the Work done by each force Work = F d cos (θ) FN Work = F d cos (θ) Fa =40N WFa = F d cos(θ) 35° WFa = (40) (8) cos(35) FF WFa = 262.13 J Fg = mg WFn = 0 J Each of these forces are perpendicular to the distance (90°), so that the works are zero. (cos 900=0): WFg = d = 8m 0 J WFf = FF d cos(θ) WFf = 3.25 (8) cos(180) We need Ff to find the work done by friction. Remember Ff =μFn WFf = -26 J Fn does not equal Fg because the Fa is at an angle. (Fa is “helping” Fn). To find Fn: Fn + (Fa sin(35)) = Fg Fn = mg - (Fa sin(35)) Fn = 16.26 N y-component of Fa Ff =μFn Ff =(0.2)(16.26) Ff =3.25 N

Work done by each force and the total work WFa = 262.13 J WFn = 0 J WFg = 0 J WFf = -26 J Total work = Wfa + WFn + WFg + WFf Total work = 262.13 + 0 + 0 + (-26) Total work = 236.13 J

Fn + Fa sinθ = Fg Ff = Fa cos θ When Fnet = 0, All forces (and components) are equal (constant velocity / no acceleration) FN Fy Fa Remember, the applied force is at an angle. There is an x and y component. θ° FF Fx Fx = Fa cos θ Fy = Fa sin θ Fg = mg Fn + Fa sinθ = Fg Ff = Fa cos θ

Work as the area under the graph Force vs Position graph Force F Positive work done Position x Negative work done

Graph of Force vs. Displacement The area under the curve is equal to the work done. Force, F F x1 x2 Work = F(x2 - x1) Area Displacement, x

What work is done by a constant force of 40 N moving a block from x = 1 m to x = 4 m? Force, F Displacement, x 1 m 4 m Area Work = F(x2 - x1) Work = (40 N)(4 m - 1 m) Work = 120 J

Mechanical Energy ME is the energy related to an objects motion or position Potential Energy (PE or Ug) Stored energy due to position No PE when object is at h=0 PE =mgh Kinetic Energy (KE or K) Energy in motion No KE if object is at rest KE= ½mv2 Elastic Potential Energy (ePE or Us) Energy stored in springs due to compression or stretching ePE= ½kx2

Kinetic and Potential Energy are EQUAL!! Total PE = Total KE There is height and velocity There is height but no velocity There is velocity Height is 0 P.E. = max K.E. = 0 P.E. = K.E. K.E. = max P.E. = 0

Energy is converted but not lost! Total Energy in this system = 100 J P.E. = 100 J K.E. = 0 P.E. = 50 J K.E. = 50 J K.E. = 100 J P.E. = 0

Elastic Potential Energy ePE is Potential energy stored as a result of deformation of an elastic object, such as the stretching of a spring.

Since we know that energy is conserved, we can make the assumption that: KE = PE = ePE = W ½ mv2 = mgh = ½ kx2 = Fd

The Work-Energy Theorem W = ΔKE Work is equal to the change in Kinetic Energy The Work-Energy Theorem: The work done by a resultant force is equal to the change in kinetic energy that it produces.

Work = ½ mvf2 - ½ mvo2 x F d = - ½ mvo2 Example 1: A 20-g projectile strikes a mud bank, going through a distance of 6 cm before stopping. Find the stopping force F if the entrance velocity is 80 m/s. x F = ? 80 m/s 6 cm Work = ½ mvf2 - ½ mvo2 F d = - ½ mvo2 F (0.06 m) = - ½ (0.02 kg)(80 m/s)2 F (0.06 m) = -64 J F = 1067 N Work to stop bullet = change in K.E. for bullet

Work = F d Work = DK 25 m FF FF = mFN = m mg DK = ½ mvf2 - ½ mvo2 Example 2: A bus slams on brakes to avoid an accident. The tread marks of the tires are 25 m long. If m = 0.7, what was the speed before applying brakes? Work = F d Work = DK 25 m FF FF = mFN = m mg DK = ½ mvf2 - ½ mvo2 Work = -m mg d -½ mvo2 = -mk mg d

Example 3: A 4-kg block slides from rest to the bottom of the 300 inclined plane. Find velocity at bottom. (h = 20 m) h 300 FN mg x

Power Power is defined as the rate at which work is done: Work (J) Force (N) Distance (m) P = ΔE = W = Fd cosθ = Fvcosθ t t t Power (watts) Time (sec) Power is measured in Watts

P = (900 kg)(9.8 m/s2)(4 m/s) cos (0°) v = 4 m/s Power Example: What power is required to lift a 900-kg elevator at a constant speed of 4 m/s? P = F v cos θ= mg v cos θ P = (900 kg)(9.8 m/s2)(4 m/s) cos (0°) P = 35300 W

Hooke’s Law When a spring is stretched, there is a restoring force that is proportional to the displacement. The larger the displacement, the larger the restoring force F x m The spring constant k is a property of the spring given by:

The stretching force is the weight (W = mg) of the 4-kg mass: Example: A 4-kg mass suspended from a spring produces a displacement of 20 cm. What is the spring constant? F 20 cm m The stretching force is the weight (W = mg) of the 4-kg mass: F = (4 kg)(9.8 m/s2) = 39.2 N Now, from Hooke’s law, the force constant k of the spring is: k = = DF Dx 39.2 N 0.2 m k = 196 N/m