D. Newton's Law of Motion: (Not Directly Stated)

REGALARIO, JEREMIAH DANIEL A.
12-STEM A
GENERAL PHYSICS 1 – PROBLEM SET #3
____________________________________________________________________________
D. Newton’s Law of Motion
D.2. Calculate the tension on the rope necessary to keep a 3 kg block from sliding down a
frictionless surface inclined at 20°.
Given: 𝑚𝑚 = 3 kg, 𝜃𝜃 = 20°, 𝑔𝑔 ⃗ = −9.8 m/s2 (not directly stated)

Asked: 𝐹𝐹𝑇𝑇⃗
Solution: Using the equilibrium formula:
� 𝐹𝐹 ⃗ = 0
� 𝐹𝐹 ⃗ = 𝐹𝐹𝑇𝑇⃗ + 𝐹𝐹𝑔𝑔⃗ sin 𝜃𝜃 = 0

𝐹𝐹𝑔𝑔⃗ sin 𝜃𝜃 = 𝑚𝑚𝑔𝑔 ⃗ sin 𝜃𝜃 = (3 kg)(−9.8 m/s2 ) sin 20° = −10.06 N
𝐹𝐹𝑇𝑇⃗ + (−10.06 N) = 0
𝐹𝐹𝑇𝑇⃗ = 10.06 N
Answer: 𝐹𝐹𝑇𝑇⃗ = 1 × 101 N

The tension force (𝐹𝐹𝑇𝑇⃗ ) necessary is 1 × 101 N (rounded to 1 significant figure)
D.4. Ethan is dragging a bag of garbage from their garage to the street corner where the
garbage collector is posted. Given the following forces acting on the bag 𝐹𝐹𝑁𝑁 ⃗ = 60 N,
⃗
𝐹𝐹𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = 20 N, 𝐹𝐹𝑓𝑓⃗ = 4 N; determine the net force on the bag and the mass and
acceleration of the garbage bag.
⃗ = 60 N,
Given: 𝐹𝐹𝑁𝑁 ⃗
𝐹𝐹𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = 20 N, 𝐹𝐹𝑓𝑓⃗ = 4 N, 𝑔𝑔 ⃗ = −9.8 m/s2
⃗ , 𝑎𝑎⃗
Asked: 𝐹𝐹𝑛𝑛𝑛𝑛𝑛𝑛
Solution: Compute for the mass: (We know 𝐹𝐹𝑁𝑁 ⃗ = −𝐹𝐹𝑔𝑔⃗ )

𝑊𝑊 𝐹𝐹𝑔𝑔⃗ 𝐹𝐹 ⃗
𝑚𝑚 = = = − 𝑁𝑁
𝑔𝑔 ⃗ 𝑔𝑔 ⃗ 𝑔𝑔 ⃗
60 N
𝑚𝑚 = − −9.8 m/s2 = 6.12 kg
Using the net force formula:

⃗ = � 𝐹𝐹 ⃗
𝐹𝐹𝑛𝑛𝑛𝑛𝑛𝑛
𝐹𝐹𝑛𝑛𝑛𝑛𝑛𝑛 ⃗ + 𝐹𝐹𝑔𝑔⃗ + 𝐹𝐹𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎

⃗ = 𝐹𝐹𝑁𝑁 ⃗ + 𝐹𝐹𝑓𝑓⃗ = 60 N + (−60 N) + 20 N + (−4 N) = 16 N
Using Newton’s formula for acceleration:

⃗
𝑎𝑎⃗ =
𝑚𝑚
16 N
𝑎𝑎⃗ = 6.12 kg
= 2.61 m/s2
⃗ = 2 × 101 N, 𝑎𝑎⃗ = 3 × 100 m/s2 (rounded to 1 significant figure).

Answer: 𝐹𝐹𝑛𝑛𝑛𝑛𝑛𝑛
D.5. AJ and Aga are members of the stage crew for a musical stage play. Between acts, they
must quickly move a grand piano onto the stage. After the curtain closes, they exert a
sudden forward force of 545 N to budge the piano from rest and get it up to speed. The 160
kg piano experiences 400 N of friction.
a. What is the piano's acceleration during this phase of its motion?

b. If AJ and Aga maintain this forward force for 1.44 seconds, then what speed will the
piano have?
Given: 𝐹𝐹𝑎𝑎⃗ = 545 N, 𝑚𝑚 = 160 kg, 𝐹𝐹𝑓𝑓⃗ = 400 N, 𝑡𝑡 = 1.44 s, 𝑣𝑣𝑖𝑖 = 0 m/s
Asked: (a) 𝑎𝑎⃗ (b) 𝑣𝑣𝑓𝑓
Solution: Compute for the net force acting on the object:

⃗ = � 𝐹𝐹 ⃗
⃗ = 𝐹𝐹𝑎𝑎⃗ + 𝐹𝐹𝑓𝑓⃗ = 545 N + (−400 N) = 145 N

Using Newton’s formula for acceleration:

⃗
𝑎𝑎⃗ =
𝑚𝑚
145 N
𝑎𝑎⃗ = = 0.91 m/s2
160 kg
Using the kinematic equation:

𝑣𝑣⃗𝑓𝑓 − 𝑣𝑣⃗𝑖𝑖
𝑎𝑎⃗ = ⟺ 𝑣𝑣⃗𝑓𝑓 = 𝑣𝑣⃗𝑖𝑖 + 𝑎𝑎⃗𝑡𝑡
𝑡𝑡
𝑣𝑣⃗𝑓𝑓 = 0 m/s + (0.91 m/s2 )(1.44 s) = 1.31 m/s
Answer:
(a) 𝑎𝑎⃗ = 1 × 100 m/s2 , (b) 𝑣𝑣𝑓𝑓 = 1 × 100 m/s . (rounded to 1 significant figure)
D.6. A 3 kg wooden toy truck is at rest on a wooden table. The coefficients of static friction and
kinetic friction are 0.25 and 0.20 respectively. If a 5.0 N horizontal force is applied on the toy
truck, what is its acceleration?
Given: 𝑚𝑚 = 3 kg, 𝜇𝜇𝑠𝑠 = 0.25, 𝜇𝜇𝑘𝑘 = 0.20, 𝐹𝐹𝑎𝑎⃗ = 5.0 N, 𝑔𝑔 ⃗ = −9.8 m/s2
Asked: 𝑎𝑎⃗
Solution: Using the equation definitions of static and kinetic friction

𝑓𝑓 𝑓𝑓
⃗ , 𝜇𝜇𝑠𝑠 = 𝑠𝑠𝑚𝑚𝑚𝑚𝑚𝑚 ⟺ 𝑓𝑓𝑠𝑠
𝜇𝜇𝑘𝑘 = 𝑘𝑘 ⟺ 𝑓𝑓𝑘𝑘 = 𝜇𝜇𝑘𝑘 × 𝐹𝐹𝑁𝑁 ⃗
= 𝜇𝜇𝑠𝑠 × 𝐹𝐹𝑁𝑁
𝐹𝐹𝑁𝑁⃗ ⃗
𝐹𝐹𝑁𝑁 𝑚𝑚𝑚𝑚𝑚𝑚
⃗ = −𝐹𝐹𝑔𝑔⃗ = −𝑚𝑚𝑔𝑔 ⃗ = −(3 kg)(−9.8 m/s2 ) = 29.4 N

𝐹𝐹𝑁𝑁
𝑓𝑓𝑘𝑘 = (0.20)(29.4 N) = 5.88 N
𝑓𝑓𝑠𝑠𝑚𝑚𝑚𝑚𝑚𝑚 = (0.25)(29.4 N) = 7.35 N
Since 𝑓𝑓𝑠𝑠𝑚𝑚𝑚𝑚𝑚𝑚 > 𝑓𝑓𝑘𝑘 > 𝐹𝐹𝑎𝑎⃗ , therefore, the wooden toy truck will remain at rest (𝑎𝑎⃗ =
2
0 m/s ) despite being applied with a force of 5.0 N. This is because the static friction can
sustain a maximum force up to 7.35 N (𝑓𝑓𝑠𝑠𝑚𝑚𝑚𝑚𝑚𝑚 ) of force before moving.
Answer: 𝑎𝑎⃗ = 0 m/s2 (rounded to 1 significant figure)

D.8. A car moving at 96.8 km/h travels around a circular curve of radius 182.9 m on a flat country
road. What must be the minimum coefficient of static friction to keep the car from slipping?
Given: 𝑣𝑣 = 96.8 km/h, 𝑟𝑟 = 182.9 m, 𝑔𝑔 ⃗ = −9.8 m/s2

Asked: min(𝜇𝜇𝑠𝑠 )
Solution: Conversion of units:

𝑣𝑣 = 96.8h km × 1000
1 km
m 1h
× 3600 s
= 26.89 m/s,
Using the net force formula for centripetal motion, we get:

2
⃗ = 𝑚𝑚 �𝑣𝑣 �
𝑟𝑟
By manipulating the equation definition of the coefficient of statis friction, we get:
𝑓𝑓𝑠𝑠
𝜇𝜇𝑠𝑠 = 𝑚𝑚𝑚𝑚𝑚𝑚 ⟺ 𝑓𝑓𝑠𝑠𝑚𝑚𝑚𝑚𝑚𝑚 = 𝜇𝜇𝑠𝑠 �𝐹𝐹𝑁𝑁 ⃗ �
⃗
𝐹𝐹𝑁𝑁
To keep the car from slipping, the static friction must be greater than or equal to the net
force acting on it:
⃗
𝑓𝑓𝑠𝑠𝑚𝑚𝑚𝑚𝑚𝑚 ≥ 𝐹𝐹𝑛𝑛𝑛𝑛𝑛𝑛
2
𝜇𝜇𝑠𝑠 �𝐹𝐹𝑁𝑁 ⃗ � ≥ 𝑚𝑚 �𝑣𝑣 �
𝑟𝑟
⃗
We know that: 𝐹𝐹𝑁𝑁 = −𝐹𝐹𝑔𝑔 = −𝑚𝑚𝑔𝑔 ⃗ ⃗
𝑣𝑣2
𝜇𝜇𝑠𝑠 (−𝑚𝑚𝑔𝑔 )⃗ ≥ 𝑚𝑚 � �
𝑟𝑟
(26.89 m/s)2
−𝜇𝜇𝑠𝑠 (−9.8 m/s2 ) ≥ � �
182.9 m
(26.89 m/s)2
� 182.9 m �
𝜇𝜇𝑠𝑠 ≥
9.8 m/s2
𝜇𝜇𝑠𝑠 ≥ 0.40
min(𝜇𝜇𝑠𝑠 ) = 0.40
Answer: The minimum coefficient of static friction (𝜇𝜇𝑠𝑠 ) needed is 4.00×10−1 . (rounded to 3
significant figures)

D. Newton's Law of Motion: (Not Directly Stated)

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REGALARIO, JEREMIAH DANIEL A.

D. Newton’s Law of Motion

Given: 𝑚𝑚 = 3 kg, 𝜃𝜃 = 20°, 𝑔𝑔 ⃗ = −9.8 m/s2 (not directly stated)

� 𝐹𝐹 ⃗ = 𝐹𝐹𝑇𝑇⃗ + 𝐹𝐹𝑔𝑔⃗ sin 𝜃𝜃 = 0

Answer: 𝐹𝐹𝑇𝑇⃗ = 1 × 101 N

Solution: Compute for the mass: (We know 𝐹𝐹𝑁𝑁 ⃗ = −𝐹𝐹𝑔𝑔⃗ )

Using the net force formula:

𝐹𝐹𝑛𝑛𝑛𝑛𝑛𝑛 ⃗ + 𝐹𝐹𝑔𝑔⃗ + 𝐹𝐹𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎

Using Newton’s formula for acceleration:

⃗ = 2 × 101 N, 𝑎𝑎⃗ = 3 × 100 m/s2 (rounded to 1 significant figure).

a. What is the piano's acceleration during this phase of its motion?

Solution: Compute for the net force acting on the object:

⃗ = 𝐹𝐹𝑎𝑎⃗ + 𝐹𝐹𝑓𝑓⃗ = 545 N + (−400 N) = 145 N

Using Newton’s formula for acceleration:

Using the kinematic equation:

Solution: Using the equation definitions of static and kinetic friction

⃗ = −𝐹𝐹𝑔𝑔⃗ = −𝑚𝑚𝑔𝑔 ⃗ = −(3 kg)(−9.8 m/s2 ) = 29.4 N

Answer: 𝑎𝑎⃗ = 0 m/s2 (rounded to 1 significant figure)

Given: 𝑣𝑣 = 96.8 km/h, 𝑟𝑟 = 182.9 m, 𝑔𝑔 ⃗ = −9.8 m/s2

Solution: Conversion of units:

Using the net force formula for centripetal motion, we get:

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