PHY 111 FINAL EXAM FINAL PAPER 2026 STUDY
GUIDE DETAILED ANSWERS READY
◉ a. Calculate the power required for a 1,400 kg car to climb a 10
degree hill at a steady 80 km/hr.
b. Calculate the power required of a 1,400-kg car to pass another car
on a level road accelerating from 90 to 110 km/hr in 6 seconds.
Assume the retarding force on the car is 700 N throughout.
Answer: a.
80 km/hr = 22.2 m/s
P = F(Vavg) = (mg sin 10o ) (22.2 m/s) = 5.289E4 W (1 hp/746 W) =
70.9 hp
b.
F= applied force
Fr= retarding force = 700 N
Vi = initial speed = 90 km/hr = 90 (5/18) = 25 m/s
Vf= final speed = 110 km/hr = 110 (5/18) = 30.6 m/s
Ax=(30.6 m/s-25.0 m/s)/6.0s=0.93 m/s^2
,MAx=net force=F-Fr
F=MAx+Fr=(1400 kg)(0.93 m/s^2)+700 N=2000N
P=Fv=(2000 N)(30.6 m/s)=6.12x10^4 Watts=82 hp 1 watt=0.00134
hp
◉ The tires of a car make 65 revolutions as the car slows down
uniformly from 100 km/hr to 50 km/hr. The diameter of the tires is
0.80 m.
a. What is the angular acceleration?
b. If the car continues to decelerate at this rate, how much more time
is required to stop the car?
Answer: a.
u = (100 km/hr)(1 m/s/3.6km/hr) = 27.78 m/s
v = (50 km/hr)(1 m/s/3.6km/hr) = 13.89 m/s
radius= diameter/2 = .80 m/2 = .40 m
v = rw
so w = v/r
wo = u/r = (27.78 m/s)/(.40 m) = 69.44 rad/s
w = v/r = (13.89 m/s)/(.40 m) = 34.72 rad/s
q = (65 rev)(2p radians/rev) = 408.41 radians
, w2 = wo2 + 2aq : wo = 69.44 rad/s; w = 34.72 rad/s; q = 408.41
radians
a = -4.4281 rad/s/s
b.
For 34.72 rad/s to zero:
wo = 34.72 rad/s
w=0
a = -4.4281 rad/s/s
w = wo + at
t = 7.841s = 7.8 s
◉ Two masses (m1 = 18.0 kg and m2 = 26. 5 kg) are connected by a
rope that hangs over a pulley. The pulley has mass = 7.5 kg and a
radius R = 0.260 m.
Initially, m1 is on the ground, and m2 rests 3.00 m above the ground.
The system is released. Use conservation of energy to determine the
speed of m2 just before it hits the ground. Assume the pulley is
frictionless
Answer: a.
m2gh = 1/2m1v2 +1/2m2v2 + m1gh + 1/2Iw^2
GUIDE DETAILED ANSWERS READY
◉ a. Calculate the power required for a 1,400 kg car to climb a 10
degree hill at a steady 80 km/hr.
b. Calculate the power required of a 1,400-kg car to pass another car
on a level road accelerating from 90 to 110 km/hr in 6 seconds.
Assume the retarding force on the car is 700 N throughout.
Answer: a.
80 km/hr = 22.2 m/s
P = F(Vavg) = (mg sin 10o ) (22.2 m/s) = 5.289E4 W (1 hp/746 W) =
70.9 hp
b.
F= applied force
Fr= retarding force = 700 N
Vi = initial speed = 90 km/hr = 90 (5/18) = 25 m/s
Vf= final speed = 110 km/hr = 110 (5/18) = 30.6 m/s
Ax=(30.6 m/s-25.0 m/s)/6.0s=0.93 m/s^2
,MAx=net force=F-Fr
F=MAx+Fr=(1400 kg)(0.93 m/s^2)+700 N=2000N
P=Fv=(2000 N)(30.6 m/s)=6.12x10^4 Watts=82 hp 1 watt=0.00134
hp
◉ The tires of a car make 65 revolutions as the car slows down
uniformly from 100 km/hr to 50 km/hr. The diameter of the tires is
0.80 m.
a. What is the angular acceleration?
b. If the car continues to decelerate at this rate, how much more time
is required to stop the car?
Answer: a.
u = (100 km/hr)(1 m/s/3.6km/hr) = 27.78 m/s
v = (50 km/hr)(1 m/s/3.6km/hr) = 13.89 m/s
radius= diameter/2 = .80 m/2 = .40 m
v = rw
so w = v/r
wo = u/r = (27.78 m/s)/(.40 m) = 69.44 rad/s
w = v/r = (13.89 m/s)/(.40 m) = 34.72 rad/s
q = (65 rev)(2p radians/rev) = 408.41 radians
, w2 = wo2 + 2aq : wo = 69.44 rad/s; w = 34.72 rad/s; q = 408.41
radians
a = -4.4281 rad/s/s
b.
For 34.72 rad/s to zero:
wo = 34.72 rad/s
w=0
a = -4.4281 rad/s/s
w = wo + at
t = 7.841s = 7.8 s
◉ Two masses (m1 = 18.0 kg and m2 = 26. 5 kg) are connected by a
rope that hangs over a pulley. The pulley has mass = 7.5 kg and a
radius R = 0.260 m.
Initially, m1 is on the ground, and m2 rests 3.00 m above the ground.
The system is released. Use conservation of energy to determine the
speed of m2 just before it hits the ground. Assume the pulley is
frictionless
Answer: a.
m2gh = 1/2m1v2 +1/2m2v2 + m1gh + 1/2Iw^2
