Physics 1307 Practice Quiz 3 Chapter 6 KE PE =KE PE ...
Physics 1307 Practice Quiz 3 Chapter 6 Problem I David jumps of a bridge with a bungee cord ( a heavy stretchable cord) tied around his ankle. He falls for 15 meters before the bungee cord begins to stretch. Dave's mass is 75 kg and we assume that the cord obeys Hook's law F=kx, with k=50 N/m. If we neglect air resistance, estimate how far below the bridge Dave will fall before coming to a stop. Ignore the mass of the cord ( not realistic, however). Problem II In a film of Jesse Owen's famous long jump in the 1936 Olympics, it is observed that his center of mass rose 1.1 m from launch point to the top of the arc. What minimum speed did he need at launch if he was also noted to be traveling at 6.5 m/s at the top of the arc?
SOLUTIONS Problem solving strategy Conservation of Energy 1. 2. 3. 4.
Draw a picture. Determine the system for which energy will be conserved: the object or objects and the forces acting. Ask yourself what quantity you are looking for, and decide what are the initial and final points. If the body under investigation changes its hight during the problem, then choose a h=0 level for gravitational potential energy. This may be chosen for convenience; the lowest point in the problem is often a good choice. 5. If springs are involved, choose the unstretched spring position to be x=0. 6. If no friction or other nonconservative forces act, then apply conservation of mechanical energy:
KE iPE i=KE f PE f
7. Solve for the unknown quantity.
8. If friction or other nonconservative forces are present, then an additional term W NC will be needed:
W NC= KE PE To be sure of which sign to give to W NC , or on which side of the equation to put it, use your intuition: is the total mechanical energy increased or decreased during the process?
1
Problem I First, a graph of the jump:
Ei
15 m
y
Ef
h=0
I have chosen the zero of potential energy at the lowest point, i.e where David comes to stop. The initial energy is given by
E i=mgh i=mg15m y
and the final energy is:
1 E f = k y2 2 equating this two relations we have:
1 mg 15m y = k y 2 2 now solving for y we obtain a quadratic equation:
−2
mg k
15m−2
mg k
y y2=0
finding (using the quadratic formula) the roots of this equation we get two possible solutions: y=40 m and y=−11 m . The negative solution is nonphysical (why?) so the distance that David drops in his fall is: h=15 m + 40 m = 55 m.
2
Problem II
Vf V0
1.1 m h=0
We chose the initial point immediately after Jessie has jumped. Thus:
1 E i = m v 20 2 and the final point when Jessie is at the highest position in the parabola. There the energy is given by:
1 E f = m v 2 f mgh f 2 using again conservation of energy we have:
1 1 m v 20= m v 2 f mgh f 2 2 solving for v 0 we get:
v 0= 6.5 m /s2 2×1.1m×9.8 m/s 2
v 0 =8 m/s
3
SOLUTIONS Problem solving strategy Conservation of Energy 1. 2. 3. 4.
Draw a picture. Determine the system for which energy will be conserved: the object or objects and the forces acting. Ask yourself what quantity you are looking for, and decide what are the initial and final points. If the body under investigation changes its hight during the problem, then choose a h=0 level for gravitational potential energy. This may be chosen for convenience; the lowest point in the problem is often a good choice. 5. If springs are involved, choose the unstretched spring position to be x=0. 6. If no friction or other nonconservative forces act, then apply conservation of mechanical energy:
KE iPE i=KE f PE f
7. Solve for the unknown quantity.
8. If friction or other nonconservative forces are present, then an additional term W NC will be needed:
W NC= KE PE To be sure of which sign to give to W NC , or on which side of the equation to put it, use your intuition: is the total mechanical energy increased or decreased during the process?
1
Problem I First, a graph of the jump:
Ei
15 m
y
Ef
h=0
I have chosen the zero of potential energy at the lowest point, i.e where David comes to stop. The initial energy is given by
E i=mgh i=mg15m y
and the final energy is:
1 E f = k y2 2 equating this two relations we have:
1 mg 15m y = k y 2 2 now solving for y we obtain a quadratic equation:
−2
mg k
15m−2
mg k
y y2=0
finding (using the quadratic formula) the roots of this equation we get two possible solutions: y=40 m and y=−11 m . The negative solution is nonphysical (why?) so the distance that David drops in his fall is: h=15 m + 40 m = 55 m.
2
Problem II
Vf V0
1.1 m h=0
We chose the initial point immediately after Jessie has jumped. Thus:
1 E i = m v 20 2 and the final point when Jessie is at the highest position in the parabola. There the energy is given by:
1 E f = m v 2 f mgh f 2 using again conservation of energy we have:
1 1 m v 20= m v 2 f mgh f 2 2 solving for v 0 we get:
v 0= 6.5 m /s2 2×1.1m×9.8 m/s 2
v 0 =8 m/s
3
