Chapter 6 - Work and Energy Flashcards
A 3 kg object moving at 10 m/s comes to rest. How much work must be done to stop it?
KEf - KE0
KE final is 0 because there is no speed
Calculate the inital KE
KE = 1/2mv^2
= 1/2(3)(10^2)
= 150 J
To stop an object, its kinetic energy needs to be 0. The work required to this is the same magnitude is as the inital KE but in the opposite direction (negative work)
so -150J
Distingish between conservative and non-conservative forces and explain gravity and friction as examples
Conservative Forces: Work done only depend on inital and final positions, not path taken. Energy lost or gained is recoverable
Gravity: lifting weight, energy is same
Nonconservative Forces: WOrk depends on path of motion
For example with friction, there is irreversible energy loss. If you only consider the inital and final position, you are not considering the work done
In this chapter, what useful information does “starts from rest” additionally provide along with the velocity?
The intial KE is 0
Outline the formulas for:
Total Work (work energy theorem)
Work done by nonconservative forces (law of conservation of mechanical energy)
Work done by conservative forces
- W = KEf - KE0
- Wnc = (KEf - KEi) + (PEf - PEi)
- Wc = PEf - PEi
A fighter jet is launched from an aircraft carrier with the aid of its own engines and a steam-powered catapult. The thrust of its engines is 2.3 × 105 N. In being launched from rest it moves through a distance of 87 m and has a kinetic energy of 4.5 × 107 J at lift-off. What is the work done on the jet by the catapult?
Wtotal = Wengines + Wcatapult
For Wtotal that is the formula KEf - KEi
KEi = 0 because launched from rest and KEf is given in the equation
For Wenegines, it’s force times displacement
A 75.0-kg skier rides a 2830-m-long lift to the top of a mountain. The lift makes an angle of 14.6° with the horizontal. What is the change in the skier’s gravitational potential energy?
DeltaPE = mgh
Pretty straightforward, but for H you have to consider the verticle component.
Hence you do
y = rsineta
y = 2830sin14.6
In what scenario is velocity exchanged in a collisions?
Perfectly elastic 1D where teh two objects collide head on and have equal mass
In Figure, a 2.0 kg package slides along a floor with speed v1 = 4.0 m/s. It then runs into and compresses a
spring until the package momentarily stops. Its path to the initially relaxed spring is frictionless, but as it compresses
the spring, a kinetic frictional force from the floor, of magnitude 20 N, acts on the package. If k = 1000 N/m, by what
distance d is the spring compressed when the package stops?
Pretty easy question actually, just recognizing that it is a work problem
Wspring + Wfriction = KEfinal - KE0
Total work = change in kinetic energy
Solve for x, it’s a quadractic equation
500x^2+20x-16
and the answer is 0.16
When is work 0? (2 options)
- If object doesn’t move even if force is applied
- When the forces are perpendicular becuase cos90 is 0, in which case in circular motion the work is always 0
Helpful hints that it is a work energy problem? (x3)
1) A moving object being stopped by a spring or friction and deceleration
2) Box being pushed up a hill/against gravity (change in potential energy)
4) Spring compressing
**if there is friction involved then it is not an energy kind of problem unless you’re given the info for the work done by friction
A student is skateboarding down a ramp that is 6.0 m long and inclined at 18° with respect to the horizontal. The initial speed of the skateboarder at the top of the ramp is 2.6 m/s. Neglect friction and find the speed at the bottom of the ramp.
Conservation of mechanical energy becuase there is no loss to friction
1/2mv^2 + mgh = 1/2mv^2
1/2(2.6)^2 + (9.8)(1.854) = 1/2(v^2)
you can cancel out m’s and you can find the height using simple trig of 6sin18 = 1.854
You should get a velocity of 6.56
