Please answer the following parts of the question below and show work for each p
ID: 1788625 • Letter: P
Question
Please answer the following parts of the question below and show work for each part.
Figure A Figure B Wall 1, Figure A shows a circular arena of radius r= 100 m with a flat floor surrounded by a frictionless vertical wall where blocks slide from To to1-Two paths between these points are labeled. The fric- tional forces from the floor and any external forces on the block-arena-Earth system may be different for each part of this problem. a) [4 pts] Suppose the floor exerts a frictional force on the blocks and external forces are absent. What force is responsible for the net centripetal force on a block taking Path 2, and what object causes it? b) [6 pts] Suppose the arena's floor has 0.25 and no external forces are present. Two blocks, both of mass m 70 kg, slide from o to1 such that their final kinetic energy is 0. lf vis the initial speed of a block taking Path 1, and v2 is the initial speed of a block taking Path 2, find v2i- V c) [4 pts] Suppose the arena's floor is frictionless and there is an external force of unknown mathemati equal is the external force conservative (Yes/No/Not Enough Information)? Justify your answer. (Hint: Consider paths other than 1 and 2.) d) [6 pts] Suppose the arena's floor exerts a frictional force on a block of unknown mass pushed at constant speed by a gladiator exerting the external force 400 N at the fixed angle (see Figure B). If the power the gladiator applies to the block is 1500 W and the gladiator completes Path 1 in 50 seconds, what is e?Explanation / Answer
for the given problem
a. the centripital force for the block to complete the path 2 in circulat direction is provided by the notmal reaction of the wall, which points inwards ( always radially inwards for a circular path)
b. given k = 0.25
m = 70 kg
hence friction force due to the floor = kmg = 0.25*9.81*70 = 171.675 N
now, work done by friction on box taking path1 = W1
W1 = kmg*(2r) = kmg*(200) = 34335 J
from work energy theorem, 0.5mv1^2 = 34335
v1 = 31.321 m/s
similiarly
W2 = kmg(pi*r) = 53933.29188 J = 0.5mv2^2
v2 = 39.254 m/s
hence
v2 - v1 = 7.93395 m/s
c. if the work done by the force on moving the block from position 1 to poisiton 2 is independent of the path( same for path1 and path2) then the force is a conservative force
d. frictional force = f
force exerted by gladiator, F = 400 N
angle the theta
power P = 1500 W
path 1 in 50 s
then
P = Fvcos(theta)
vcos(theta) = P/F = 3.75 m/s
also,
v = 2*r/50 = 200/50 = 4 m/s
hence
theta = 20.3641 deg