When work is done by a completely unbalanced force it can set objects in motion…or increase their speed. You push and it simply coasts away beyond your reach.

d F d F F d When an object moves in the direction of an unbalanced push it accelerates over the entire distance d. W = F  d produces the acceleration a = F/m accelerating through the distance d in a time t: d = ½ at 2

Work is done on a wagon, initially at rest, by pushing it with an unbalanced force F over a distance D. With what final speed does it coast away? Final speed? With what acceleration a does it build speed? But for how long does it accelerate? So finally

W = F  d But there’s another way to think about this F = mad = at W = ma  at 2 So the amount of work done is also equal to: 1212 W = ma 2 t Then, just recall that the acceleration builds up, over t, to a final velocity: 0, if starting from rest! W = mv We consider this an expression of the energy possessed by a mass m when moving with a velocity v

Moving objects carry energy to scatter pins to knock down walls In fact, energy is defined as the ability to perform work! push snow aside

A piledriver uses a raised weight to drive piles into the ground. Raised weights also possess energy …potential energy… the potential to fall and build kinetic energy.

Raising a mass against gravity is doing work, even if there is no resulting final velocity.

Work done in lifting a mass to a height h: W = F  d average force lifted with mg W = mgh In falling through a distance h a mass builds speed to what final kinetic energy? And in that time reaches KE = mgh So:

Weight Normal Force If tipped completely of course the ramp gives NO SUPPORT! A horizontal surface (if strong enough to hold up!) can give full support to an object’s weight What about any of the positions in between?

Weight Normal Force Normal Force The unbalanced (net) force is parallel to the inclined surface of the road! Exactly the direction we know a car left in neutral will start to coast! We can “add” these forces in a diagram by showing the arrows that represent them, flowing from head into tail from one to the other (in either order in fact!)

hd Pushing this box up along the ramp is 1. more work 2. less work 3. exactly the same work as simply lifting it straight up h. x d W N F WdWd = F?F? 1. h2. N3. x

hd Work pushing box along the ramp is Work in simply lifting the box h x d W N F WdWd = FhFh Force in lifting  distance lifted W  h F  dF  d

A man, his briefcase in hand as shown, walks down the corridor from elevator to office door. He does 1. positive work on the briefcase. 2. no work on the briefcase. 3. negative work on the briefcase. F d

A man, his briefcase in hand as shown, walks down the corridor from elevator to office door. He does 1. positive work on the briefcase. 2. no work on the briefcase. 3. negative work on the briefcase. F d The force F does is not along (nor back against) the direction of motion. Instead F and d are perpendicular! The force of support neither raises nor lowers the briefcase (neither increasing or decreasing its potential energy). It Neither speeds up nor slows down the briefcase (neither increasing nor decreasing the briefcases kinetic energy).

h d x d W N F WdWd = FhFh The box may slip down from its position on the shelf by sliding back down the ramp or falling straight down to the floor. By the time it reaches the floor it will reach 1. a higher final speed by falling. 2. a higher final speed by sliding. 3. the same final speed by either route. assume frictionless rollers on ramp

h dd W N F WdWd = FhFh Falling: Sliding: then

Some Answers 3. exactly the same work Question 1 This won’t be obvious until we work through the next several questions! 1. h Question 2 The small triangle drawn to display the forces acting other box, and the large triangle formed by the ramp, wall and floor are similar. d is the hypotenuse of the large triangle, W the hypotenuse of the small one. 2. no work on the briefcase. Question 3 The force F does is not along (nor back against) the direction of motion. Instead F and d are perpendicular! The force of support neither raises nor lowers the briefcase (neither increasing or decreasing its potential energy). It neither speeds up nor slows down the briefcase (neither increasing nor decreasing the briefcases kinetic energy). 3. the same final speed by either route. Question 4 The small triangle drawn to display the forces acting other box, and the large triangle formed by the ramp, wall and floor are similar. d is the hypotenuse of the large triangle, W the hypotenuse of the small one. But already this much should be clear: lifting straight up requires a large force. Pushing along the ramp a smaller force, but the push must be applied over a much larger distance (the full length of the ramp). As you’ll see by the end of this lecture the proportions in the triangles guarantee the two compensate exactly! The large force time small distance = the smaller force times longer distance.