Other
PHYS204 Physics Assignment 2
- Course
- PHYS204 (PHYS204)
- Institution
- Athabasca University (AU )
Physics 204 Assignment 2 covering Chapters 5 - 7, the laws of motions, newton's laws, circular motion, and energy of a system
[Show more]
Connected book
Book Title:
Author(s):
- Edition:
- ISBN:
- Edition:
Seller
Follow
Content preview
Athabasca University PHYS 204 Assignment 2Assignment 2
The ten questions below cover Units 5–7 and each carries equal weight. You are expected to solve all
questions in detail and submit by uploading to the appropriate drop box. Scanned copy of your
handwritten solutions should be acceptable.
1. The system below is released from rest and moves 90 cm in 1.5 s, without friction. What is the mass
m of the hanging object?
Finding acceleration first and net force:
1
d = v1 t + 2 a xt 2
1
0.90 m = (0)(1.5 s) + 2 a x (1.5)2
a x = 0.80 m/s2
⃑⃑⃑⃑ = ma⃑
ΣF
m
⃑⃑⃑⃑ = (2.0 kg) (0.80 ) = 1.6 N
ΣF
s2
Treating the box like a particle under a net force model:
x-component of box
ΣFx = mg
⃑T = m1 a⃑x
The y-component:
ΣFy = 0
(no movement in the y-direction)
Treating the ball like a particle under a net force model:
The x-component:
ΣFx = 0
(no movement in the x-direction)
The y-component:
ΣFy = mg
⃑ = m2 a⃑
m2 g − T
Solving for T and substituting in the other equation:
, ⃑T = m1 a⃑x
m2 g − ⃑T = m2 a⃑
m2 g − m1 a⃑x = m2 a⃑
m 𝑚 m
m2 (9.80665 2 ) − (2.0 kg)(0.8 2 ) = m2 (0.8 2 )
s 𝑠 s
m m 𝑚
m2 (9.80665 2 ) − m2 (0.8 2 ) = (2.0 kg)(0.8 2 )
s s 𝑠
m2 (9.80665 − 0.8) = 1.6 N
1.6
m2 = = 0.177646 kg
9.00665
The mass of the hanging object is 1.78 × 10−1 kg.
2. When the two boxes shown below are released from rest the system accelerates with
2.80 m/s2. Calculate the coefficient of kinetic friction, assuming that it is the same for both
boxes.
Normal force
tension force Mass 2
F = ma 70
tension force 𝑚𝑔𝑠𝑖𝑛20
70
Normal force 20
Gravitational force
𝑚𝑔 cos 20
Gravitational force
𝑚𝑔 cos 60
90
F = ma
𝑚𝑔 sin 60
For Mass 1 (2.5 kg):
⃑ = m1 g⃑ cos 60
n
m
⃑ = (2.5 kg) (9.80665
n ) cos 60
s2
⃑ = 12.2583 N
n
⃑ = m1 a⃑
F
m1 g⃑ sin 𝜃 − ⃑T − fk = m1 a⃑
m1 g⃑ sin 𝜃 − ⃑T − (n)(μ) = m1 a⃑
m
(2.5 kg)(9.80665)(sin 60) − ⃑T − (12.2583)(μ) = (2.5 kg) (2.80 2 )
s
−T ⃑ = 7.0 − 21.232 + 12.2583μ
⃑T = 14.232 − 12.2583μ
⃑F 7.0 N
μ= = = 0.571042
mg⃑ 12.2583 N
For Mass 2 (4.0 kg):
The benefits of buying summaries with Stuvia:
Guaranteed quality through customer reviews
Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.
Quick and easy check-out
You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.
Focus on what matters
Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!
Frequently asked questions
What do I get when I buy this document?
You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.
Satisfaction guarantee: how does it work?
Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.
Who am I buying these notes from?
Stuvia is a marketplace, so you are not buying this document from us, but from seller bronzeandromeda2016. Stuvia facilitates payment to the seller.
Will I be stuck with a subscription?
No, you only buy these notes for CA$14.39. You're not tied to anything after your purchase.
Can Stuvia be trusted?
4.6 stars on Google & Trustpilot (+1000 reviews)
72042 documents were sold in the last 30 days
Founded in 2010, the go-to place to buy study notes for 14 years now