Physics Study Guide: Work, Energy, Forces, and Motion

Work and Energy

Work Done by a Force

Work is defined as the product of the force applied to an object and the displacement of the object in the direction of the force.

• Formula: $W = F imes d imes ext{cos}( heta)$, where $F$ is the force, $d$ is the displacement, and $heta$ is the angle between the force and displacement vectors.

• Example: A 20 N horizontal force moves a 4.0 kg box 3.0 meters. If friction is negligible, $W = 20 imes 3.0 = 60$ J.

Gravitational Potential Energy

Gravitational potential energy is the energy stored in an object due to its position in a gravitational field.

• Formula: $ext{PE}_g = mgh$, where $m$ is mass, $g$ is acceleration due to gravity, and $h$ is height.

• Example: A 30.0 kg box slides up a 12.0 m incline at 30° above the horizontal. Change in height $h = 12.0 imes ext{sin}(30^ ext{o}) = 6.0$ m. $ext{PE}_g = 30.0 imes 9.8 imes 6.0 = 1,764$ J.

Kinetic Energy and Work-Energy Principle

Kinetic energy is the energy of motion. The work-energy principle states that the net work done on an object is equal to its change in kinetic energy.

• Formula: $ext{KE} = rac{1}{2}mv^2$

• Work-Energy Principle: $W_{ ext{net}} = ext{KE}_{ ext{final}} - ext{KE}_{ ext{initial}}$

• Example: A 120 N force moves a 4.0 kg box 6.0 m. The change in kinetic energy is $W = 120 imes 6.0 = 720$ J.

Forces and Motion

Average Force from Bullet Penetration

The average force exerted on an object can be found using the work-energy theorem, relating the work done to the change in kinetic energy.

• Formula: $F_{ ext{avg}} = rac{ ext{KE}_{ ext{initial}}}{d}$, where $d$ is the penetration distance.

• Example: A 5.00 g bullet at 300 m/s penetrates 0.10 m. $ext{KE} = rac{1}{2} imes 0.005 imes 300^2 = 225$ J. $F_{ ext{avg}} = rac{225}{0.10} = 2,250$ N.

Inclined Plane and Friction

When a force is applied to move an object up an incline, both the component of the force along the incline and friction must be considered.

• Kinetic Friction Formula: $f_k = ext{coefficient} imes N$, where $N$ is the normal force.

• Example: A 20.0 kg box slides up a 12.0 m incline at 30°, with a force of 150 N applied at 10° above the incline. If the coefficient of kinetic friction is 0.100, the increase in kinetic energy is calculated by subtracting work done against friction and gravity from the total work.

Velocity on a Slope

The velocity of an object sliding down a frictionless slope can be found using energy conservation.

• Formula: $v = ext{sqrt}(2gh)$

• Example: A 75.0 kg skier slides down a 75.0 m high slope. $v = ext{sqrt}(2 imes 9.8 imes 75.0) = 38.4$ m/s.

Spring Force and Energy

Work Done by a Spring

The work done by a spring is given by the change in elastic potential energy.

• Formula: $W = rac{1}{2}k(x_f^2 - x_i^2)$, where $k$ is the spring constant, $x_i$ and $x_f$ are initial and final compressions.

• Example: Compressing a spring with $k = 8.0$ N/m from 5.0 cm to 20.0 cm: $W = rac{1}{2} imes 8.0 imes (0.20^2 - 0.05^2) = 0.15$ J.

Projectile from a Spring Gun

The maximum height of a projectile shot from a spring gun can be found using energy conservation and projectile motion equations.

• Formula: $h_{ ext{max}} = rac{v_0^2 ext{sin}^2 heta}{2g}$, where $v_0$ is initial velocity, $heta$ is launch angle.

• Example: A spring-powered dart gun with $k = 150$ N/m, compressed 8.0 cm, launches a 5.0 g projectile at 30°. Calculate $v_0$ from spring energy, then use projectile formula for $h_{ ext{max}}$.

Power

Power Needed to Run Up Stairs

Power is the rate at which work is done or energy is transferred.

• Formula: $P = rac{W}{t}$, where $W$ is work and $t$ is time.

• Example: A 90.0 kg person runs up stairs at 0.500 m/s. Power needed is $P = mgh/t = mgv = 90.0 imes 9.8 imes 0.5 = 441$ W.

Circular Motion and Tension

Tension in a Pendulum String

At the bottom of a swing, the tension in the string must support both the weight of the bob and provide the centripetal force for circular motion.

• Formula: $T = mg + rac{mv^2}{r}$, where $m$ is mass, $v$ is velocity, $r$ is length of string.

• Example: A 2.00 kg pendulum bob on a 1.50 m string, released from 60° with $v = 0$ at release. Find $v$ at bottom using energy conservation, then calculate $T$.

Summary Table: Key Formulas

Concept	Formula	Units
Work	$W = Fd ext{cos} heta$	Joules (J)
Gravitational Potential Energy	$ext{PE}_g = mgh$	Joules (J)
Kinetic Energy	$ext{KE} = rac{1}{2}mv^2$	Joules (J)
Spring Work	$W = rac{1}{2}k(x_f^2 - x_i^2)$	Joules (J)
Power	$P = rac{W}{t}$	Watts (W)
Tension (Pendulum)	$T = mg + rac{mv^2}{r}$	Newtons (N)

Additional info: These notes expand on the original questions by providing definitions, formulas, and example calculations for each concept, making them suitable for exam preparation in a college-level Physics course.