**What is work?**

Work is done when there is a force that acts on a moving system in the direction of the displacement.

*work is a scalar, but can be positive or negative.

*Units of work – Nm, or Joules (J)

▪ Positive Work – A force does positive work if a component of the force acts in the direction of displacement.

▪ Negative Work – A force does negative work if a component of the force acts opposite the direction of displacement.

▪ Zero Work – A force does zero work if the force is perpendicular to displacement, so that no component of force is in direction of displacement.

*If several forces act on an object, the total work is the algebraic sum of the individual works.

**Example A**: A mass is pulled up an incline that is a distance 𝑑 long by a force F that acts parallel to the plane. The coefficient of sliding friction between the mass and the incline is 𝜇.

a) What is the work done by the pull force F?

**Answer/Explanation**

Ans: F does positive work since it points in the direction of motion:

𝑊 = 𝐹𝑑 cos 0 = +𝑭𝒅

b) What is the work done by gravity?

**Answer/Explanation**

Ans: Gravity does negative work since a component of it (mgsin𝜃) opposes the direction of motion.

𝑊 = (𝑚𝑔 sin 𝜃)𝑑 cos 180° → −𝒎𝒈 𝐬𝐢𝐧 𝜽 𝒅

c) What is the work done the normal force?

**Answer/Explanation**

Ans: The normal force does zero work since it is perpendicular to the direction of motion.

d) What is the work done by friction?

**Answer/Explanation**

Ans: Friction does negative work since is opposite the direction of motion.

𝑊 = 𝐹_{𝑓}𝑑 cos 180° = −𝝁𝒎𝒈 𝐜𝐨𝐬 𝜽𝒅

e) What is the total work done on the mass?

**Answer/Explanation**

Ans: 𝑾 = 𝑭𝒅 − 𝒎𝒈 𝐬𝐢𝐧 𝜽 𝒅 − 𝝁𝒎𝒈 𝐜𝐨𝐬 𝜽𝒅

Calculate work done by a varying force.

▪ If the force applied varies with distance: \(W = \int_{a}^{b}F(x)dx\)

▪ If the force applied varies in direction and magnitude: \(W = \int_{a}^{b}F(x)cos \theta dx\)

**Example B**: A force varies with distance as given below:

𝐹(𝑥) = 6𝑥^{2} + 5 sin 𝑥.

**Answer/Explanation**

Ans: The force is applied to a 3.0 kg object that is initially at rest with no other forces acting. Find the speed of the object after it has been pushed 5 meters.

\(W = \int_{2}^{5}F(-4x^{3})dx=-609 Nm\)

Work to stretch a spring

▪ Review: Hooke’s Law (Force to stretch a spring):

Hooke’s Law relates what distance from equilibrium (x) a spring will be displaced when a force F is applied to stretch or compress the spring.

𝑭 = −𝒌𝒙

k – spring constant (units: N/m) *the negative sign indicates the force is opposite displacement

▪ Work to stretch a spring: 𝑾 = ∫ 𝑭 𝒅𝒙

**Example B**: A grocery market uses a non–ideal spring with a force given by \(F = -\frac{1}{2}kx^{3}9.0 kg\) of bananas stretch the spring scale 0.3 m.

What is the spring constant of the scale?

**Answer/Explanation**

Ans: The spring force counter-acts the weight of the bananas:

*it is against societal norms to buy so many bananas

How much would work would it take to stretch the spring .5 m?

\(W = \int_{0}^{.5}(\frac{1}{2}kx^{3})dx=250 j\)

**What is the work energy theorem?**

Objective: Apply the work–energy theorem.

▪ Kinetic Energy – A scalar quantity that measures the energy of an object’s motion.

𝑲 = 𝟏/𝟐𝒎𝒗^{𝟐}

▪ The Work Energy Theorem – The (net) work done by an external force on an object is equal to the change in kinetic energy of the object.

Doing work on a system increases (positive work) or decreases (negative work) that the energy the system has. The net work (by nonconservative forces) is equal to the change in kinetic energy for a system in motion.

W=∆𝐾

➢ *Doing work on a system increases the energy of a system. When a system does negative work, it loses energy.

▪ If force varies with position:

𝑊 = ∫ 𝐹𝑑𝑥 = ∆𝐾**Example A**: A 1.0 kg block is attached to a spring (k=3.0 N/m) and rests on a rough horizontal surface. The spring is initially uncompressed. The block is given an initial velocity of 10 m/s and compresses a maximum distance of 4.0 m. What is the coefficient of sliding friction?

**Answer/Explanation**

Ans:

Work done by friction and the spring change the block’s kinetic energy as it compresses the spring.

𝑊 = ∆𝐾