Work, Kinetic Energy, and Power

v f 2 = v i 2 + 2ad and F = ma v f 2 -v i 2 = 2ad and F/m = a v f 2 -v i 2 = 2(F/m)d Fd = ½ mv f 2 – ½ mv i 2 Fd = Work (W) ½ mv 2 = Kinetic Energy (KE) A Constant Force Acts on the Backpack Through a Distance

Work (W) Product of the net force applied to an object and the displacement of the object in the direction of the applied force. –No work is done on an object if the net applied force does not cause motion (Ex: Holding a bag of groceries). –Work is done when a force acts parallel to the displacement or has a component parallel to the displacement.

Kinetic Energy (KE) The energy that an object has due to its motion. KE = ½ m v 2 –KE  m and KE  v 2 –Kinetic energy is a scalar quantity. F d v i = 0

Work-Energy Theorem Fd = ½ mv f 2 – ½ mv i 2 If Fd =W, W = ½ mv f 2 – ½ mv i 2 If KE = ½ mv 2, W = KE f – KE i W = ∆ KE (Work-Energy Theorem) –Energy units are the same as work units (kg*m 2 /s 2 ) = N*m = Joules (J)

Calculating Work W = Fd (cos  ) F = Applied force (N) d = displacement of object (m)  = angle between F and d W = Work done –Units = N*m = Joules (J) 1 J = 1 N*m 1 J = 1 N*m F  d

Positive and Negative Work Positive work means that the object’s kinetic energy increases. Negative work means that the object’s kinetic energy decreases. –Forces that do negative work have a component in the opposite direction of the displacement (  =180 o )

Ex: How much work is done on an 8.0 kg wagon rolling on a flat sidewalk if a force of 60. N is used to stop the wagon in a distance of 15 m? Given: F = 60. N m = 8.0 kg m = 8.0 kg d = 15 m d = 15 m  = 180 o  = 180 o Find: W = ? W = Fd cos  W = (60.N)(15 m) (cos 180 o ) W = -9.0 x 10 2 J

Ex: How much work is done in lifting a 12 kg crate from the floor to a platform 3.0 m above the floor? Given: m = 12 kg d = 3.0 m d = 3.0 m Find: W = ? W = F d cos  = (mg) d cos  = (12 kg)(9.80 m/s 2 )(3.0 m) (cos 0 o ) = 3.5 x 10 2 J

Power (P) The rate at which work is done or energy is transferred. Power is calculated by dividing the work done by the time required to do the work. P = W/  t P= (Fd cos  ) /  t Units: = J/s = N  m /s = watts (W)

Power (P) Because d/t = v, power can also be calculated: P = Fv Another unit of power is horsepower (hp). –1 hp = 746 W Doubling the power output means that the same amount of work can be done in half the time, or that twice the work can be done in the same amount of time.

Ex: A horse pulling a cart exerts a force of 250. N over a distance of 100. m for 20.0 s. What is the power used by the horse in watts and in horsepower? Given: F = 250. N d = 100. m Δt = 20.0 s Find: P = ? P = W/t P = F (cosθ) d / Δt = 250. N (cos 0)(100.m) / 20.0 s = 1250 W 1250 W (1hp / 746 W) =1.68 hp