Answers to posted exam 1:
a) 75.1 N
b) 26.3 N
c) 10a = 63.0 - T - 26.3
d) 2a = T - 19.6
e) a = 1.43 m/s
2
; T = 22.5 N
(A) magnitude approx. = 7; about 25 deg. west of north
a) x
hobo
= 3t
b) x
train
= t + (1/2)(0.4)t
2
Hint: Solution is at x
hobo
= x
train
+ 4.
c) Yes; t = 2.77 s.
d) 2.11 m/s
a) v
0x
= 12.99 m/s; v
0y
= 7.5 s.
b) H = 26.2 m
c) D = 41.6 m
a) 6432 J
b) V = 12.7 m/s
c) W
f
= -6432 J
d) Coeff. of friction = 0.41