Work Done on an Object

• Work occurs when a force is applied to an object, causing it to move from one location to another
• Work describes the amount of energy transferred when an object is displaced due to an applied force, this can include pulling, pushing, or lifting an object

Work Done on an Object

• Work is the product of the magnitude of displacement and the component of force parallel to the displacement
• The equation for work is W = F||d, where F|| is the component of force parallel to displacement d
• The equation for work can also be expressed as W = Fdcosθ, where F is the magnitude of the force, d is the magnitude of the displacement, and θ is the angle between the force and displacement vectors
• The unit for work is the Joule (J), equivalent to a Newton-meter (N·m)

Signs of Work

• Work is positive if the force component points in the same direction as the displacement vector
• Work is negative if the force component points in the opposite direction of the displacement vector
• Work is zero if the force component is perpendicular to the displacement

Example Problems: Work

• The work done by the weight lifter on the barbell in the lifting phase is 460 J
• The work done by the weight lifter during the lowering phase is -460 J
• To determine the work done by each force acting on the sled, calculate the work done by friction: WFk=−70,000 J≈−70 kJ
• To determine the work done by each force acting on the sled, calculate the work done by tension: WFT=80,000 J≈80 kJ
• The total work done by all the forces: WT=10,000 J≈10 kJ
• The magnitude of the force the shopper exerts is 54.9 N
• The work done by the pushing force is 1,056 J
• The work done by the frictional force is -1,056 J
• The work done by the gravitational force is 0 J

Seatwork Problems & Solutions: Work

• 1. Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.80 x 10^6 N, one 14° west of north and the other 14° east of north, as they pull the tanker 0.75 km toward the north. The total work they do on the supertanker is 2.62 x 10^9 J
• 2. A factory worker pushes a 30.0 kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and the floor is 0.25.
• a. What magnitude of force must the worker apply? Answer:73.5 J
• b. How much work is done on the crate by this force? Answer: 330.75 J
• c. How much work is done on the crate by friction? Answer: -330.75 J
• d. How much work is done on the crate by the normal force? By gravity? Answer: 0J
• e. What is the total work done on the crate? Answer: 0 J
• 3. Suppose the worker in the previous item pushes downward at an angle of 30° below the horizontal:
• a. The magnitude of force the worker must apply to move the crate at a constant velocity: 99.2 N
• b. How much work is done on the crate by this force when the crate is pushed a distance of 4.5 m? Answer: 387 J
• c. How much work is done on the crate by friction during this displacement? Answer: 387 J approx.
• d. How much work is done on the crate by the normal force? By gravity? Answer: 0 J, 0 J
• e. What is the total work done on the crate? Answer: 0 J, no energy transferred
• 4. A tow truck pulls a car 5.00 km along a horizontal roadway using a cable having a tension of 850 N:
• a. Work done by the cable on the car if it pulls horizontally: 425 kJ
• a. Work done by the cable on the car if it pulls at 35 degrees above the horizontal: 348 KJ
• b. Work done the cable does on the tow truck in both cases of part (a): -425 KJ, -348 KJ
• c. Work does gravity do on the car in part (a): 0 J

Work-Energy Theorem

• Newton's second law explains a net force accelerates objects, and this acceleration changes the magnitude of velocity or speed
• W = (1/2)mv2^2 -(1/2)mv1^2, where (1/2)mv^2 quantifies the kinetic energy of an object
• KE = (1/2)mv^2
• The SI unit of kinetic energy is the Joule (J)
• Kinetic energy is energy in motion that can be zero if the object is at rest but can never be negative
• The Work-Energy Theorem states: W = KEf - KEi which is equal to W = (change in)KE

Implications of the Work-Energy Theorem

• Positive work increases an object's kinetic energy
• Negative work decreases an object's kinetic energy
• If negative, the applied net force on the object is opposite in the direction of motion

Example Problems: Work-Energy Theorem

• Determining the speed of an object after it moves 20 m: 4.2 m/s
• What is the car's mass to accelerate from 23.0 m/s to 28.0 m/s.: 1450 kg
• To determine the speed of the arrow as it leaves the bow: 43 m/s

Conservation of Energy

• Potential energy is the energy that is stored in an object associated with its position relative to a reference point (position zero); it measures the potential of work to be done
• Gravitational Potential Energy (PEgrav) is the energy stored in an object due to gravitational interaction with Earth, related to the object's vertical position
• PEgrav = mgh, where g is gravitational acceleration, m is the mass, and h is the height from a reference point

Conservation of Mechanical Energy

• In the absence of dissipative forces, the total mechanical energy of the system is conserved; the sum of initial kinetic energy (KEi) and initial potential energy (PEi) equals the sum of final kinetic energy (KEf) and final potential energy (PEf)
• PEi + KEi = PEf + KEf

Elastic Potential Energy

• Elastic potential energy is the energy stored in an elastic object (like a spring) due to its deformation (stretching or compression)
• This energy depends on the length of compression/expansion (x) and the spring constant (k), which indicates stiffness
• PEelastic = (1/2)kx^2

Power

• Power (P) is defined as the rate at which energy is produced, transferred, or transformed, expressed as
• P = Energy / Δt where energy can be the form of work done or change in KE or PE and it measured in J/s
• The SI unit of power is the Watt (W), where 1 W = 1 J/s
• Another common larger unit of power is horsepower (hp), and 1 hp = 746 W

Example Power Word Problems

• Two weightlifters lifting barbells of equal mass to a specific height at varying rates, assuming weightlifter (a) transfers energy more gradually while weightlifter (b) transfers energy more rapidly
• Weightlifter (a) produces less power than weightlifter (b)
• To find total energy consumed for a given time frame: Energy = P∆t

Momentum

• Linear momentum describes an object's momentum in translational motion representing the inertia of a moving object
• Mathematically: p = mv, where p is momentum, m is mass, and v is velocity
• The SI unit for momentum is kg m/s

Properties of Momentum

• Momentum is a vector quantity
• Momentum can be decomposed into its components
• Horizontal momentum: px = mvx
• Vertical momentum: py = mvy
• Momentum is related to kinetic energy

Equations Between Momentum & Kinetic Energy

• KE= p^2/2m

Impulse

• The net force required to change the motion of an object equals the time rate of the change in momentum: Fnet = Δp / Δt
• The impulse is the net force times the amount of time of that force: J = Fnet Δt
• The SI unit is Newton-second (Ns)

Impulse Momentum Theorem

• The impulse of the net force on a body during a time interval is equal to the change in momentum of that particle during the