To solve letter a:
d1 = 85t1 = 16 km,
85t1 = 16,
t1 = 16 / 85 = 0.1882 h = 11.29 min.
d2 = 115t2 = 16 km,
115t2 = 16,
t2 = 16 / 115 = 0.139 h = 8.35 min.
t1 - t2 = 11.29 - 8.35 = 2.94 min.
Car #2 arrives 2.94 minutes sooner.
To solve letter b:
15 min = 1/4 h = 0.25 h.
d1 = d2,
115t = 85(t + 0.25),
115t = 85t + 21.25,
115t - 85t = 21.25,
30t = 21.25,
t = 21.25 / 30 = 0.71 h,
d = 115 * 0.71 = 81.65 km.
Answer:
(a) 81.54 N
(b) 570.75 J
(c) - 570.75 J
(d) 0 J, 0 J
(e) 0 J
Explanation:
mass of crate, m = 32 kg
distance, s = 7 m
coefficient of friction = 0.26
(a) As it is moving with constant velocity so the force applied is equal to the friction force.
F = 0.26 x m x g = 0.26 x 32 x 9.8 = 81.54 N
(b) The work done on the crate
W = F x s = 81.54 x 7 = 570.75 J
(c) Work done by the friction
W' = - W = - 570.75 J
(d) Work done by the normal force
W'' = m g cos 90 = 0 J
Work done by the gravity
Wg = m g cos 90 = 0 J
(e) The total work done is
Wnet = W + W' + W'' + Wg = 570.75 - 570.75 + 0 = 0 J
Average speed = Distance traveled / time taken
In this case Time taken = Difference in hours between 3 PM and 9 AM
= 6 hours
Total distance traveled = 350 km + 250 km
= 600 kilometers
So average speed = 600/6 = 100 km/hr
Average speed of bus = 27.78 m/s
So the bus's average speed = 27.78 m/s or 100 km/hr.
