PHY121Ch 6-9 ExamName

You are a member of an alpine rescue team and must get a 3.00 kg box of supplies up a 30° incline so that it reaches a stranded skier who is a vertical distance 4.00 m above the bottom of the incline. There is some friction present, µ = 0.100. Since you can't walk up the incline, you give the box a push that gives it an initial velocity; then the box slides up the incline, slowing down under the forces of friction and gravity. To aid in the ascent, you attach one of your bottle rockets to the back side of the box of supplies, as shown, which exerts a force of 10.0 N. Use the work-energy theorem to calculate the minimum speed v that you must give the box at the bottom of the incline so that it will reach the skier. /
TEbottom= TEtop
[KEpush + PEbottom] + FR-in dir d – Ff d = KEtop + PEtop
KEpush + PEbottom + FR-in dir d – Ff d = 0 + mghtop
½3v2 + 0 + 10 cos30° 8 - µ FN8= m g htop
½3v2 + 0 + 69.3 - µ FN8= 3 10 (4)
5/53/3 5/5 5/5 5/5 5/5 5/5
1.5v2 + 69.3 - (0.1)(10sin30 + mgcos30) 8= 120 J
v = 7.1 m/s / Note, please account for energy contributed by the bottle rocket, FR d in the initial energy, similar but opposite to the non-conservative force, Friction through a distance, Ff d.
sin θ = h / d
sin 30 = 4 / d
d = 8 meters 1/1
A 50.0 kg roller-coaster car is released from rest and slides along a frictionless track. At ground level, the car encounters a loop of radius R = 10 m. If the track exerts a normal force of 625 N on the car at the top of the loop, what is the initial height of the car?
Both mg and the normal force are toward the center, thus additive. /
2/2 8/8 2/2 7/7
mg + 700 = mv2 / r
50(10) + 625 = 50 v2 / 10
50(10) + 625 = 50 v2 / 10
v =15 m/s / TEtoptrack= TEtoploop
4/4 2/2 8/8
mghtrack + ½mvtrack2= mghloop + ½mvloop2
500(htrack) + 0= 500(2R) + ½50 152
htrack = 31.25 m
A 50.0 kg roller-coaster car is moving 4 m/s at h = 30 meters on the top of the track and slides along a frictionless track. At ground level, the car encounters a loop of radius, R. If the track exerts a normal force of 200 N on the car at the top of the loop, what is R? / OR for 80% max 
/ A 50.0 kg roller-coaster car is moving 10 m/s at h = 30 meters on the top of the track and slides along a frictionless track. At ground level, the car encounters a loop of radius, R = 10 m. What is the normal force exerted by the track on the car at the top of the loop?
2/2 8/8 2/2 7/7
mg + 200 = mv2 / r
50(10) + 200 = 50 v2 / R
14 R = vtoploop2 / Both mg and the normal force are toward the center, thus additive. / TEtoptrack= TEtoploop
4/4 2/2 2/2 4/4
mghtrack + ½mvtrack2= mghloop + ½mvloop2
500(30) + ½50 102= 500(2R) + ½50 vloop2
vloop = 17.3 m/s
TEtoptrack= TEtoploop
4/4 2/2 2/2 6/6
mghtrack + ½mvtrack2= mghloop + ½mvloop2
500(30) + ½50 42= 500(2R) + ½50 14R
15400= 1350R
R = 11.4 meters / 2/2 6/6 2/2 4/4
mg + FN = m v2 / r
50(10) + FN = 50 17.32 / 10
FN = 1000 N

A 0.500 kg block rests on a frictionless table and is attached to a horizontal spring (k = 30 N/m) that is uncompressed. A 0.100 kg wad of putty is launched horizontally on the frictionless table by another identical spring which is compressed by 10 cm. The wad impacts the blocks and sticks. How far does the putty-block system compress the original spring?

Workspring = ΔK
5/5 4/4
½ k x2 = ½ m v2
½30(.1)2 = ½(.1)vwad2
vwad = 1.73 m/s / po = pf
7/7 7/7
mvblock + mvwad = (mblock + mwad) vf
0 + .1 (1.73) = (0.5 + 0.1) vf
vf = 0.289 m/s / Workspring = ΔK
5/5 5/5
½ k x2 = ½(mblock + mwad) vf2
½30x2 = ½(mblock + mwad) vf2
x = 4.08 cm

The collision between a sledge hammer and a nail for a rail tie can be considered to be approximately elastic. Calculate the kinetic energy acquired by a 500-g nail when it is struck by a 8 kg hammer moving with an initial speed of 5.0m/s.

8/8 8/8
mh vh + mn vn = mh vhf + mn vnf
8 (5) + ½ (0) = 8vhf + ½vnf
80 - 16vhf = vnf / 4/4 4/4 4/4
½mh vh2 + ½mn vn2 = ½ mh vhf2 + ½mn vnf2
½8 52 + ½ ½ 02 = ½ 8 vhf2 + ½ ½ vnf2
400 - 16 vhf2 = vnf2
400 - 16 vhf2 = (80 - 16 vhf)2
Vhf = 4.41 m/s; Vnf = 9.44 m/s / 5/5
Ko = 0
Kf = ½ m v2
Kf = ½ ½ 9.442
Kf = 22.3 J