Transcript 101KnightChp12_7.pptx

Textbook - Knight
Chapter 10
(page 312)
Chap 10:9
• A car is travelling 10 m/s.
• a. How fast would the car need to go to double its
kinetic energy?
–
KE = ½mv² = ½m(10)² = 50m
Double the KE would be 100m:
–
KE = ½m(v2)² = 100m
–
cancel m and multiply by 2:
–
v2 = √200 = 14.1 m/s
(v2)² = 200
• b. By what factor does a car’s kinetic energy
increase if its speed is doubled.
– Old KE
= ½mv²
– If we change v to 2v:
KE = ½m(2v)² = ½m4v² = 2mv²
– Ratio = (2 mv²) / (½mv²) = 2 / ½ = 4
Chap 10:16
• The lowest point in Death Valley is 85 meters
below sea level. The summit of nearby Mt.
Whitney has an elevation of 4420 meters. What is
the change in gravitational potential energy of a
65 kg hiker who makes it from Death Valley to the
top of Mt. Whitney?
• Δ PE = mgh
•
= PE2 − PE1 = mgy2 − mgy1
•
= mg(y2−y1) = 65 ⋅ 10 ⋅ (4420 −(−85))
•
= 65 ⋅ 10 ⋅ (4505) = 2,928,250 joules
Chap 10:24
• Marissa drags a 23 kg duffel bag 14 meters across
a gym floor. If the coefficient of friction between
the floor and the bag is .15, how much thermal
energy does Marissa create?
• Ffriction = μsFN = .15 x 230 = 34.5 newtons
• Work done by friction = Ffriction ⋅ distance
•
= 34.5 newtons x 14 meters
•
= 483 joules
•
All of this energy turns into heat.