A. True friction is two objects touching each other
Explanation:
here u = 50m/s
v = 60m/s
t = 58 s
then a = (60-50)/58 m/s2
= 0.17m/s2
now s= ut+1/2at2
so , 50×58+0.5×0.17×(58)^2 m
= 3185.94 m
= 3.18 km
Answer:
(a) the high of a hill that car can coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h is 47.6 m
(b) thermal energy was generated by friction is 1.88 x
J
(C) the average force of friction if the hill has a slope 2.5º above the horizontal is 373 N
Explanation:
given information:
m = 750 kg
initial velocity,
= 110 km/h = 110 x 1000/3600 = 30.6 m/s
initial height,
= 22 m
slope, θ = 2.5°
(a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h?
according to conservation-energy
EP = EK
mgh = 
gh = 
h = 
= 47.6 m
(b) If, in actuality, a 750-kg car with an initial speed of 110 km/h is observed to coast up a hill to a height 22.0 m above its starting point, how much thermal energy was generated by friction?
thermal energy = mgΔh
= mg (h -
)
= 750 x 9.8 x (47.6 - 22)
= 188160 Joule
= 1.88 x
J
(c) What is the average force of friction if the hill has a slope 2.5º above the horizontal?
f d = mgΔh
f = mgΔh / d,
where h = d sin θ, d = h/sinθ
therefore
f = (mgΔh) / (h/sinθ)
= 1.88 x
/(22/sin 2.5°)
= 373 N
T = tension force in the rope in upward direction
m = mass of the box attached at end of rope = 56 kg
W = weight of the box in downward direction due to gravity
a = acceleration of the box in upward direction = 5.10 m/s²
weight of the box is given as
W = mg
inserting the values
W = (56) (9.8)
W = 548.8 N
force equation for the motion of the box is given as
T - W = ma
inserting the values
T - 548.8 = (56) (5.10)
T = 834.4 N
Using the formula F=ma
500N=50kg (a)
a= 10 m/s^2