Physics Notes on Work, Energy and Power

$W = Fd \cos\theta$
$W = \vec{F} \cdot \vec{d}$
- Where $F \cos\theta$ is the component of the force parallel to the displacement $d$ .
SI Unit of work is joule (J)
- $1 J = 1 N.m$

Movers pushing a 160 kg crate 10.3 m across a rough floor without acceleration, with $\mu_k = 0.50$
Free body diagram shows forces acting on the crate.
Net force in the x-direction: $\sum Fx = F - fk = ma = 0$
Therefore, $F = f_k$
Friction force: $fk = \muk N = \mu_k mg$
Work done: $W = Fd \cos\theta = \mu_k mgd \cos(0^{\circ})$
$W = 0.50 \times 160 \times 9.8 \times 10.3 \times \cos(0^{\circ}) = 8.1 \times 10^3 J$

Minimum work needed to push a 950 kg car 810 m up a 9.0° incline.
- (a) ignore friction
- (b) include friction, $\mu_k = 0.25$
Free body diagram shows forces acting on the car on an incline.
Net force in the x-direction: $\sum F_x = F - mg \sin\theta = 0$
Net force in the y-direction: $\sum F_y = N - mg \cos\theta = 0$
Minimum work occurs when the car is moved with constant velocity.
(a) When friction is ignored:
- $W = Fd = mg \sin\theta \times d$
- $W = 950 \times 9.8 \times \sin(9.0^{\circ}) \times 810 m = 1.2 \times 10^6 J$
(b) When friction is included:
- $\sum Fx = F - mg \sin\theta - fk = 0$
- $F = mg \sin\theta + f_k$
- $fk = \muk N = \mu_k mg \cos\theta$
- $F = mg \sin\theta + \mu_k mg \cos\theta$
- $W = Fd = (mg \sin\theta + \mu_k mg \cos\theta)d$
- $W = (950 \times 9.8 \times \sin(9.0^{\circ}) + 0.25 \times \cos(9.0^{\circ})) \times 810 m = 3.0 \times 10^6 J$
Note: Work in part (b) is more than part (a), as expected, because some of the work is done to overcome friction.

Energy: The ability to do work.
Kinetic Energy (KE):
- The energy of an object by virtue of its motion.
- $KE = \frac{1}{2} mv^2$ (translational KE only)
Potential Energy (PE):
- The energy of an object by virtue of its position or configuration.
- Gravitational Potential Energy (GPE): $mgh$
- Elastic Potential Energy: $\frac{1}{2} kx^2$
Translational motion: objects moving without rotation

A 1500 kg car traveling at 17 m/s can brake to a stop within a distance of 20 m.
How much work is done by the friction force?
- $W = \Delta K = Kf - Ki$
- $W = \frac{1}{2} mvf^2 - \frac{1}{2} mvi^2$
- $W = 0 - \frac{1}{2} \times 1500 \times (17)^2 = -2.2 \times 10^5 J$
The direction of the friction force: opposite to the direction of motion.

A person accelerates a 50 kg box from 0.5 m/s to 2.5 m/s over a distance of 2.0 m.
- (a) How much work is done by the person?
- (b) How much force is applied by the person? Assume no friction.
(a) Work done by the person:
- $W = \Delta K = Kf - Ki$
- $Ki = \frac{1}{2} mvi^2$
- $Kf = \frac{1}{2} mvf^2$
- $W = \frac{1}{2} \times 50 \times (2.5)^2 - \frac{1}{2} \times 50 \times (0.5)^2$
- $W = 1.5 \times 10^2 J$
(b) Force applied by the person:
- $W = Fd \cos\theta$
- $F = \frac{W}{d \cos\theta} = \frac{1.5 \times 10^2}{2.0 \times \cos(0^{\circ})} = 75.1 N$

What will be the final speed if another child with twice the mass slides down?
- Final speed of both children will be the same, as mass cancels out in the CME equation.
Will the final speed be greater or smaller if friction is present?
- Smaller, as some of the energy is used up in overcoming friction.

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The time rate of doing work is called power.
- $P = \frac{W}{t}$
- The average power due to the force is:
  - $P = \frac{F \cdot d}{t}$
  - $P = F \cdot v$

How long will it take a 1750 W motor to lift a 285 kg piano to a sixth-story window 16.0 m above?
- $P = \frac{W}{t} = \frac{mgh}{t}$
- $t = \frac{mgh}{P} = \frac{285 \times 9.8 \times 16.0}{1750} = 25.5 s$

A shot putter accelerates a 7.3 kg shot from rest to 14 m/s. If this motion takes 1.5 s, what average power was developed?
- $P = \frac{\Delta K}{t} = \frac{\frac{1}{2} mvf^2 - \frac{1}{2} mvi^2}{t}$
- $P = \frac{\frac{1}{2} \times 7.3 \times (14)^2}{1.5} = 4.8 \times 10^2 W$