Conservation Laws Work

Definitions and equations for work Energy Method Force Method Work Kinetic Energy Theorem Work is equal to a change in kinetic energy. Changes in kinetic arise from equal changes in potential energy. However, when kinetic energy increases and is positive, potential energy decreases and is negative (and vice versa). As a result the theorem is generalized to state that work is a change in energy. Work is a force through a distance Work is equal to the dot product of the force and displacement vectors. d is the linear displacement of the object and can be replaced with any variable representing displacement (Δx , Δy , Δh , etc.). The “dot” in a vector dot product is essentially an abbreviation for cosine.

Sign conventions Energy Method Force Method If the object’s speed increases, then the kinetic energy increases. If the object’s speed decreases, then the kinetic energy decreases. If the object’s speed is constant, then kinetic energy is constant. Rearrange the formula slightly. F cos θ is the component of force parallel to displacement, d . Work is positive when F|| and d point in the same direction. Work is negative when F|| and d point in opposite directions.

Solving for specific types of work Energy Method Force Method Energy equations can be changed into work equations by enclosing the equation in parenthesis and then precede it with a delta symbol. For potential energies include a minus sign. The work of any force can be calculated. Determine the component of force that is parallel to displacement or the component of displacement that is parallel to the force. Moving down Moving up Friction opposes motion Net work needs the net force, ΣF

Units of work F Δr θ

Three Key scenarios 1st θ = 0o F d m A mass m is acted upon by a 20 N force and moves a distance of 10 m. The force acts in the direction of motion. Determine the work done on the mass. θ = 0o Solving with the full formula, including cos θ Solving using parallel force, like last chapter Force is parallel to motion and is in direction of motion (positive).

Three Key scenarios 2nd θ = 180o F d m A mass m is acted upon by a 20 N force and moves a distance of 10 m. The force acts opposite the direction of motion. Determine the work done on the mass. θ = 180o Solving with the full formula, including cos θ Solving using parallel force, like last chapter Force is parallel to motion and is opposite of motion (negative).

Three Key scenarios 3rd θ = 90o F d m A mass m is acted upon by a 20 N force and moves a distance of 10 m. The force acts perpendicular to the motion. Determine the work done on the mass. θ = 90o F Solving with the full formula, including cos θ Solving using parallel force, like last chapter Force is perpendicular to motion. Forces perpendicular to motion do no work.

Three Key Scenarios summarized F d θ = 90o F d θ = 0o F d θ = 180o Positive F Positive a Positive W Negative F Negative a Negative W F has no effect in direction of motion a = 0 in direction of motion W = 0 Increases speed Gains energy Decreases speed Losses energy Constant speed in direction of initial motion

Example 1 Need the force applied (A) Using W = F d A 2.0 kg mass is moving at 20 m/s when it is acted upon by a force in the direction of motion. The force acts over a distance of 30 m, and it increases the speed of the mass to a final speed of 40 m/s. Determine the work done during the period that the force acted (A) Using W = F d Need the force applied (B) Now try Work Kinetic Energy Theorem Often using energy is faster and easier than force and kinematics.

Example 2 A 30 N force, F , is applied at an angle of 37o above the horizontal to a 10 kg mass, as shown. The mass moves 5.0 m at a constant velocity along a rough horizontal surface. F θ m d (A) Determine the work done by the applied force, F . (B) Determine the work done by friction. Moving at constant velocity At constant velocity friction has the same magnitude, but opposite direction as the component of force pulling parallel to the surface.

Example 2 F θ m d (C) Determine the work of gravity. A 30 N force, F , is applied at an angle of 37o above the horizontal to a 10 kg mass, as shown. The mass moves 5.0 m at a constant velocity along a rough horizontal surface. F θ m d (C) Determine the work of gravity. Perpendicular forces do no work. (D) Determine the net work. At constant velocity the sum of forces (net force) is zero.

Example 3 A force, F , is applied to a 3.0 kg mass pushing it a distance, d = 4.0 m up a 30o frictionless incline at constant velocity. (A) Determine the change in height, Δh . θ = 37o d = 4.0 m Δh = ? (B) Determine the change in potential energy of the mass, ΔU .

Example 3 A force, F , is applied to a 3.0 kg mass pushing it a distance, d = 4.0 m up a 30o frictionless incline at constant velocity. (C) Determine the work of gravity, Wg , done by the force of gravity, Fg . θ −Fg sin θ +d −Fg +Δh Moving upward Δh = +2.0 m

Example 3 A force, F , is applied to a 3.0 kg mass pushing it a distance, d = 4.0 m up a 30o frictionless incline at constant velocity. (D) Determine the work, W , of the applied force, F . The applied force, F , is opposite and Equal to the force of gravity. θ −Fg sin θ +d −Fg +Δh The applied force is not associated with potential energy, and acts opposite gravity, changing the minus sign to positive. +F +Fy When an applied force does work against gravity the magnitude of work is the same as the work of gravity, but the sign is reversed.

Example 3 FORCE METHOD ENERGY METHOD A force, F , is applied to a 3.0 kg mass pushing it a distance, d = 4.0 m up a 30o frictionless incline at constant velocity. (E) Determine the net work, Wnet . The net work can be solved two ways: FORCE METHOD ENERGY METHOD Use the net force At constant velocity a = 0 & ΣF = 0 Use Work Kinetic Energy Theorem At constant velocity Δv = 0 & ΔK = 0

Example 3 A force, F , is applied to a 3.0 kg mass pushing it a distance, d = 4.0 m up a 30o frictionless incline at constant velocity. The 3.0 kg mass is now released from rest and slides 4.0 m along the frictionless 30o incline. θ = 37o d = 4.0 m Δh = 2 m Fg sin θ (F) Determine the change in potential energy of the mass, ΔU .

Example 3 The 3.0 kg mass is now released from rest and slides 4.0 m along the frictionless 30o incline. (G) Determine the work of gravity, Wg , done by the force of gravity, Fg . θ +Fg sin θ +d +Fg +Δh Moving downward Δh = −2.0 m

Example 3 FORCE METHOD ENERGY METHOD The 3.0 kg mass is now released from rest and slides 4.0 m along the frictionless 30o incline. (E) Determine the net work, Wnet . The net work can be solved two ways: FORCE METHOD ENERGY METHOD Use the net force Since this is the same force as used in the previous slide, the answer will be the same. Use Work Kinetic Energy Theorem Kinematics would be needed to find the speed at the end of the problem, in order to solve

Work of Gravity (or against gravity) The work of gravity is only affected by changes in height. In fact if you ever see a change in height in any problem all year long, think mgΔh Why does only height matter to the work of gravity ??? Gravity is a vertical force, and only vertical motion is parallel to it resulting in work. Any motion perpendicular to the force of gravity (horizontal motion of any kind) does no work of gravity. The four masses at the left all move the same vertical distance Δh and the same work mgΔh is done on all of them.

Sign Confusion OK, I get the formulas, but the signs seem too convoluted. Yes they are, so here is a way to keep the sign of work simple. The work of a specific force (Wg , Ws , etc.) Solve the math (mgΔh , etc.) without worrying about the sign. Then examine force and displacement vectors: If they have the same direction, then work is positive. If they have opposite direction, then work is negative. What about net work? If an object speeds up, then net work is positive. If an object slows down, then net work in negative. If an object has constant speed, then net work is zero.

Example 4 A 3.0 kg mass is attached to a spring, with spring constant 200 N/m. A force, F , is applied to the mass causing it o move to the right while stretching the spring 20 cm. (A) Determine the change in potential energy. Δx F Fs (B) Determine the work of the spring, Ws , due to the restoring force, Fs . The restoring force, Fs , opposes displacement and does negative work. Solve the NEGATIVE of the change in potential energy.

Example 4 A 3.0 kg mass is attached to a spring, with spring constant 200 N/m. A force, F , is applied to the mass causing it o move to the right while stretching the spring 20 cm. (C) Determine the work, W , done by the applied force, F . Δx F Fs The force, F , needed to displace a spring is equal and opposite the restoring force, Fs . Therefore, the work of the applied force will also be equal to the work of the spring, but opposite (POSITIVE). (D) Determine the net work, Wnet , done while moving the mass 20 cm right. As long as the spring starts at rest and ends at rest, the forces acting on the spring while it is stretched are equal and opposite.

Example 4 A 3.0 kg mass is attached to a spring, with spring constant 200 N/m. A force, F , is applied to the mass causing it o move to the right while stretching the spring 20 cm. The spring/mass system is released from rest (The applied force is gone, and the direction of displacement is reversed.) Δx Fs (E) Determine the change in potential energy. (F) Determine the work of the spring, Ws , due to the restoring force, Fs . The restoring force, Fs , is in the same direction as displacement. (G) Determine the net work, Wnet , done on the mass during the 20 cm motion.

Example 5 A 2.0 kg object slides 5.0 m along a rough surface, μk = 0.5 . (A) Determine the work of friction if the surface is horizontal. N Fg Not moving in y-direction f (B) Determine the work of friction if the surface is inclined at 30o . Normal force on an incline

Work Graphing work F When force is graphed height = F When force is graphed from the initial position, d0 , to the final position, d , the height of the graph is equal to F , and the base is equal to d. Work d0= 0 base = d d The product of height base is area.

Constant vs. Variable forces Constant Force Gravity F d Work 20 N 5 m Variable Force Springs F d 20 N 5 m Work

Zero Net work Non zero Net work Zero net work occurs when objects are stationary moving at constant speed in uniform circular motion (constant speed) Non zero Net work Which force should be used in W = F d to calculate the same value for work that would be obtained using W = ΔK ? Work Energy Theorem solves for the net work, and this requires the sum of forces, ΣF .