Work, Energy and Power

Week 1

PHYSICS 11 - General Physics 2

Work is applied when a body undergoes displacement upon application (or removal) of Force.

A body moves with Force ( ) acts on the same direction as the displacement ( ).

Work

F

s

)(cos sFifFsWFsW

yFxFWsFW yx

Determine the work done if a box is pushed with a 210 N force, given its displacement.

If force applied with an angle of 53.1°, determine the work done for the same set up beforehand.

A steady force F, given in unit vector notation below is used to push a car resulting to a displacement s.

Work done by a constant force

210 N

18 m

jmims

jNiNF

ˆ)13(ˆ)7(

ˆ)30(ˆ)180(

energy transferred to or from an object by means of a force acting on the object.

Energy transferred to the object is positive work, and energy transferred from the object is negative work.

Work

A. An elevator is being lifted: The cable does positive work on the elevator, and the elevator does negative work on the cable.

B. An elevator is being lowered: The cable does negative work on the elevator, and the elevator does positive work on the cable.

Work: Positive, Negative or Zero

Solve for the total work, given the problem above:

Total Work

F=35 N

friction = 25 N

sFWT

θ = 53.1°

Kinetic Energy

Kinematics equations (for a = constant)

Newton’s Second Law of Motion

savv

atvv

attvs

x2

2

1

2

1

2

2

12

2

0

amF

energy associated with the state of motion of an object,

the faster the object moves, the greater is its kinetic energy.

Kinetic Energy

𝐾 =1

2𝑚𝑣2

𝑚 − mass of the object

𝑣 − velocity of the object

unit: 1 joule = 1 J = 1kg ∙ 𝑚2

𝑠2

For a particle, a change Δ𝐾𝐸 in the kinetic energy equals the net work 𝑊 done on the particle:

Δ𝐾𝐸 = 𝐾𝐸𝑓 − 𝐾𝐸𝑖 = 𝑊

Which can be rearranged to:

𝐾𝐸𝑓 = 𝐾𝐸𝑖 +𝑊

Work – Kinetic Energy Theorem

A sled with a mass of 1500 kg is pulled with a horizontal force equal to 1,400 N. It travels with an initial speed equal to 2.5 m/s. What is it’s speed after it moves 10 m?

Calculate speed using work and energy

Calculating Work (Calculus approach)

𝑊 = 𝐹𝑥𝑑𝑥𝑥2

𝑥1

W = 𝐹𝑎𝑥Δ𝑥𝑎 + 𝐹𝑏𝑥Δ𝑥𝑏 +⋯

(Varying 𝑥 – component of force, straight-line

displacement.)

Constant force, 𝐹 in the 𝑥-direction.

Calculating Work

𝑊 = 𝐹𝑠 = 𝐹 𝑥2 − 𝑥1

Work done by gravitational force 𝐹𝑔:

𝑊𝑔 = 𝐹𝑔𝑑

𝑊𝑔 = 𝑚𝑔𝑑 cos 𝜙

Variable Force , 𝐹 in the 𝑥-direction.

Calculating Work (Calculus approach)

𝑊 = 𝐹 𝑥 𝑑𝑥𝑥𝑓

𝑥𝑖

Work done by spring force: 𝐹𝑠 = −𝑘𝑥 (Hooke’s Law)

𝑊𝑠 = 𝐹𝑠𝑑𝑥𝑥𝑓

𝑥𝑖

22

2

1ifs xxkW

The power due to a force is the rate at which that force does work on an object. If the force does work 𝑊 during a time interval t, the average power due to the force over that time interval is:

𝑃 =𝑑𝑊

𝑑𝑇, 𝑃𝑎𝑣𝑔 =

𝑊

Δ𝑡

Power

1. (a) In the Bohr model of the atom, the ground-state electron in hydrogen has an orbital speed of 2190 km/s. What is its kinetic energy?

(b) If you drop a 1.0-kg weight (about 2 lb) from a height of 1.0 m, how many joules of kinetic energy will it have when it reaches the ground? (c) Is it reasonable that a 30-kg child could run fast enough to have 100 J of kinetic energy?

Sample Problems: Kinetic Energy

2.a. Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.80 × 106N one 14° west of north and the other 14° east of north, as they pull the tanker 0.75 km toward the north. What is the total work they do on the supertanker?

2.b. A loaded grocery cart is rolling across a parking lot in a strong wind. You apply a constant force 𝐹 = 30𝑁 𝑖 −40𝑁 𝑗 to the cart as it undergoes a displacement 𝑠 = −9.0𝑚 𝑖 − 3.0𝑚 𝑗 How much work does the force you apply do on the grocery cart?

Sample Problems: Total Work

3. A boxed 10.0-kg computer monitor is dragged by friction 5.50 m up along the moving surface of a conveyor belt inclined at an angle of 36.9° above the horizontal. If the monitor’s speed is a constant 2.10 cm/s, how much work is done on the monitor by (a) friction, (b) gravity, and (c) the normal force of the conveyor belt?

Sample Problems: Work due to gravity

4. To stretch a spring 3.00 cm from its unstretched length, 12.0 J of work must be done. (a) What is the force constant of this spring? (b) What magnitude force is needed to stretch the spring 3.00 cm from its unstretched length? (c) How much work must be done to compress this spring 4.00 cm from its unstretched length, and what force is needed to compress it this distance?

Sample Problems: Work due to spring force

5. A little red wagon with mass 7.00 kg moves in a straight line on a frictionless horizontal surface. It has an initial speed of 4.00 m/s and then is pushed 3.0 m in the direction of the initial velocity by a force with a magnitude of 10.0 N. (a) Use the work–energy theorem to calculate the wagon’s final speed. (b) Calculate the acceleration produced by the force. Use this acceleration in the kinematic relationships formula to calculate the wagon’s final speed. Compare this result to that calculated in part (a).

Sample Problems: Work – Energy Theorem