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HOW TO SOLVE PHYSICS PROBLEMS
THE PROCEDURE
• Step 1:Draw a free body diagram
• Step 2:Write down the givens• Step 3:Write down the
unknown• Step 4:Resolve the free body
diagram
THE PROCEDURE
• Step 5:Use equations for the x and y directions
• Step 6:Use “sum of” substitution• Step 7:Plug in givens and solve
for unknown• Step 8:Check the answer with
common sense
Problem 11 (page 128)
Two forces, F1 and F2, act on a 5.00 kg block. Two forces, F1 and F2, act on a 5.00 kg block. The magnitudes of the Forces are F1=45.0 N The magnitudes of the Forces are F1=45.0 N and F2=25.0 N. What is the horizontal and F2=25.0 N. What is the horizontal acceleration of the block?acceleration of the block?
F1
F265
Problem 11 (page 128)Step 1: Step 1: Draw free Draw free body diagram.body diagram.
F265
F1
+y
x
Problem 11 (page 128)
• m = 5.00 kg
• F1 = 45.0 N
• F2 = 25.0 N
Step 2: Step 2: Write Write down the givens.down the givens.
Problem 11 (page 128)
• Horizontal Acceleration = ax
Step 3: Step 3: Write down the Write down the unknown.unknown.
Problem 11 (page 128)Step 4: Step 4: Resolve the Resolve the
free body diagram.free body diagram.
F265
F1
+y
x
F2x
F2y
Problem 11 (page 128)Step 5: Step 5: Use equations for Use equations for
the x and y directions.the x and y directions.F2
65
F1
+y
x
∑Fx=ma
∑ Fy=ma
Problem 11 (page 128)
F1x - F2 = max
F1y = may
Step 6: Step 6: Use ”sum of” Use ”sum of” substitution.substitution.
F1 cos 65 -F2 = ax
mF1 cos 65 -F2 = ax
m
Problem 11 (page 128)Step 7: Step 7: Plug in Plug in
givens and solve.givens and solve.
Problem 11 (page 128)
-1.20 m/s
Step 8: Step 8: Check with Check with common sense.common sense.
2
Sounds good!Sounds good!
Problem 55 (page 131)
A helicopter is traveling horizontally to the A helicopter is traveling horizontally to the right at constant velocity. The Weight of the right at constant velocity. The Weight of the helicopter is 53800 N. The lift force L helicopter is 53800 N. The lift force L generated by the rotating blade makes an generated by the rotating blade makes an angle of 21 degrees with respect to the angle of 21 degrees with respect to the verticle. What is the magnitude of the lift verticle. What is the magnitude of the lift force and determine the magnitude of the air force and determine the magnitude of the air resistance R that opposes the motion.resistance R that opposes the motion.
Problem 55 (page 131)Step 1: Step 1: Draw free Draw free body diagram.body diagram.
L
R
W
21
+y
x
Problem 55 (page 131)
W = 53800 N
Step 2: Step 2: Write Write down the givens.down the givens.
Problem 55 (page 131)
The lift force (L)
Air resistance (R)
Step 3: Step 3: Write down Write down the unknown.the unknown.
Problem 55 (page 131) Step 4: Step 4: Resolve the Resolve the
free body diagram. free body diagram.
L
R
W
21
+y
x Lx
Ly
Problem 55 (page 131) Step 5: Step 5: Use equations Use equations
for the x and y for the x and y directions.directions.
R W
21
+y
x Lx
Ly
∑Fx = ma
∑ Fy = ma
Problem 55 (page 131)
Lx - R = ma
Ly - W = ma
Step 6: Step 6: Use ”sum of” Use ”sum of” substitution.substitution.
W = L
(cos 21)
Problem 55 (page 131) Step 7: Step 7: Plug in Plug in
givens and solve.givens and solve.
L sin 21 = R
Problem 55 (page 131)
L = 57627 N
R = 20651 N
Step 8: Step 8: Check with Check with common sense.common sense.
Problem 79 (page 133 )
The weight of a block on a table is 111 N and The weight of a block on a table is 111 N and that of the hanging block is 258 N. Assuming that of the hanging block is 258 N. Assuming the coefficient between the 111 N block and the coefficient between the 111 N block and the table is 0.300, find the acceleration of the the table is 0.300, find the acceleration of the blocks and the tension of the blocks and the tension of the string.string.
Problem 79 (page 133 )Step 1: Step 1: Draw free Draw free body diagram.body diagram.
+N
W1
F TT
W2
Problem 79 (page 133 )
u =0.300
W1 = 111 N
W2 = 258 N
Step 2: Step 2: Write Write down the givens.down the givens.
Problem 79 (page 133 )
Acceleration of the two blocks (a)
Tension of the string (T)
Step 3: Step 3: Write down Write down the unknown.the unknown.
Problem 79 (page 133 ) Step 4: Step 4: Resolve the Resolve the
free body diagram. free body diagram.+
N
W1
F TT
W2
Problem 79 (page 133 ) Step 5: Step 5: Use Use
equations for the x equations for the x and y directions.and y directions.
+N
W1
F
T
T
W2
∑Fx = ma
∑ Fy = 0∑ Fx = ma
Problem 79 (page 133 )
T-F = ma
W1-N = 0
W2-T = ma
Step 6: Step 6: Use ”sum of” Use ”sum of” substitution.substitution.
T = m1(a + ug)
a = W2-T m2
Problem 79 (page 133 ) Step 7: Step 7: Plug in Plug in
givens and solve.givens and solve.
Problem 79 (page 133 )
T = 100.89 N
a = 5.974 m/s
Step 8: Step 8: Check with Check with common sense.common sense.
2
Problem 111 (page 135)A 45 kg box is sliding up an incline that makes A 45 kg box is sliding up an incline that makes an angle of 15 degrees with respect to the an angle of 15 degrees with respect to the horizontal. The coefficient of friction between horizontal. The coefficient of friction between the box and the surface of the incline is 0.180. the box and the surface of the incline is 0.180. The initial speed of the box at the bottom of The initial speed of the box at the bottom of the incline is 1.5 m/s, How far does the box the incline is 1.5 m/s, How far does the box travel along the incline before coming to rest. travel along the incline before coming to rest.
Problem 111 (page 135)Step 1: Step 1: Draw free Draw free body diagram.body diagram.
15
N
FW
+
Problem 111 (page 135)
m = 45 kg
u = 0.180
Vo = 1.5 m/s
Step 2: Step 2: Write Write down the givens.down the givens.
Problem 111 (page 135)
Distance the box travels (d)
Step 3: Step 3: Write down Write down the unknown.the unknown.
Problem 111 (page 135) Step 4: Step 4: Resolve the Resolve the
free body diagram.free body diagram.N
FW
Wx
Wy
15
Problem 111 (page 135) Step 5: Step 5: Use equations for Use equations for
the x and y directions.the x and y directions.N
FW
Wx
Wy
15
∑Fx = ma
∑ Fy = 0
Problem 111 (page 135)
- F - Wx = ma
N - Wy = 0
Step 6: Step 6: Use ”sum of” Use ”sum of” substitution.substitution.
a = -4.245 m/s
d = V - Vo
2a
Problem 111 (page 135) Step 7: Step 7: Plug in Plug in
givens and solve.givens and solve.2
22
Problem 111 (page 135) Step 8: Step 8: Check with Check with
common sense.common sense.
d = 0.265 m
by ANDREW MEDLEYby ANDREW MEDLEY
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