Problem No. 3Two blocks are connected by a massless rope as shown below. The mass of the block on the table is 4.0 kg and the hanging mass is 1.0 kg. The table and the pulley are frictionless.Newton's second law:ΣF=maUniform accelerated motion (UAM) equations, a.k.a. "kinematics equations": vf = v0 -gt∆y= (vf+v02)t∆y= v0t-12gt2 vf2= v02 -2g∆y(a)Applying Newton's second law to the block on the table:ΣF = maT = m1a1 Applying Newton's second law to the hanging block: Tension pulls up, weight is directed downward, and the acceleration is directed downward.ΣF = maT - m2g = - m2a2m1a1 - m2g = - m2a2 m1a1 + m2a2 = m2ga1 = a2 = a(m1 + m2)a = mga = m2g/(m1 + m2) = (1 × 9.8)/(4.0 + 1.0) = 9.8/5 = 1.96 m/s2 (b) The acceleration of the system is 1.96 m/s2 QUESTIONAn elevator weighs 1960 N and is being supported by a cable having atension equal to 2300 N. What is the acceleration of the elevator? f FacebookA Physics November 10 Act - Goog xA MODULE 5 PROBLEM SET->A docs.google.com/document/d/1aREpjEqEtccCsqeKD561vXfHQHHFZNZAG9wvNzcnww0/edit?fbclid=lwAR3Hzj3imvprWw7PYONz-ZKpdd-bD-4u59kRiOPCiKjZi. QPhysics November 10 ActA ShareFile Edit View Insert Format Tools Add-ons HelpLast edit was made on December 11, 2021 by CANACO JENRY KIMI UAE : : = E : E - E - E E2 Editing100% -Normal textArialB(c) if a constant force just enough to start the crate moving is applied toit, what would be its acceleration?- Problem No.27. An elevator weighs 1960 N and is being supported by a cable havingatension equal to 2300 N. What is the acceleration of the elevator?9:50 PMO Type here to search2 25°C Mostly clear1/2/2022 f Facebookx B Physics November 10 Act - Goog xA MODULE 5 PROBLEM SETx +A docs.google.com/document/d/1aREpjEqEtccCsqeKD561vXfHQHHFZNZAG9wvNzcnww0/edit?fbclid=lwAR3Hzj3imvprWw7PYONz-ZKpdd-bD-4u59kRIOPCiKjZi. QPhysics November 10 Act * e OA ShareFile Edit View Insert Format Tools Add-ons HelpLast edit was made on December 11, 2021 by CANACO JENRY KIM14 + BI U AE E = = 1EE E X100% - Normal textArial= E - IE2 EditingProblem No. 3- Problem No.2Two blocks are connected by a massless rope as shown below. Themass of the block on the table is 4.0 kg and the hanging mass is 1.0 kgThe table and the pulley are frictionless.Newton's second law:ΣFEmaUniform accelerated motion (UAM) equations, a.k.a. "kinematicsequations":vf = v0 -gtay= (vf+v02)tay= vot-12gt2 vf2= v02 -2gay(a)Applying Newton's second law to the block on the table:EF = maT= m,a,Applying Newton's second law to the hanging block: Tension pulls up,weight is directed downward, and the acceleration is directed downward.EF = maT- m,g = - m,azm,a, - m,g = - m,azm,a, + m,a, = m,gа, - а, - а(m, + m2)a = mga = m2g/(m1 + m2) = (1 x 9.8)/(4.0 + 1.0) = 9.8/5 = 1.96 m/s2(b) The acceleration of the system is 1.96 m/s29:50 PMO Type here to search2 25°C Mostly clear1/2/2022

Answered: QUESTION An elevator weighs 1960 N and…

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Problem No. 3

Two blocks are connected by a massless rope as shown below. The mass of the block on the table is 4.0 kg and the hanging mass is 1.0 kg. The table and the pulley are frictionless.

Newton's second law:

ΣF=ma

Uniform accelerated motion (UAM) equations, a.k.a. "kinematics equations":

vf = v0 -gt∆y= (vf+v02)t∆y= v0t-12gt2 vf2= v02 -2g∆y

(a)

Applying Newton's second law to the block on the table:

ΣF = ma

T = m1a1

Applying Newton's second law to the hanging block: Tension pulls up, weight is directed downward, and the acceleration is directed downward.

ΣF = ma

T - m2g = - m2a2

m1a1 - m2g = - m2a2

m1a1 + m2a2 = m2g

a1 = a2 = a

(m1 + m2)a = mg

a = m2g/(m1 + m2) = (1 × 9.8)/(4.0 + 1.0) = 9.8/5 = 1.96 m/s2

(b) The acceleration of the system is 1.96 m/s2

QUESTION

An elevator weighs 1960 N and is being supported by a cable having a

tension equal to 2300 N. What is the acceleration of the elevator?

f Facebook
A Physics November 10 Act - Goog x
A MODULE 5 PROBLEM SET
->
A docs.google.com/document/d/1aREpjEqEtccCsqeKD561vXfHQHHFZNZAG9wvNzcnww0/edit?fbclid=lwAR3Hzj3imvprWw7PYONz-ZKpdd-bD-4u59kRiOPCiKjZi. Q
Physics November 10 Act
A Share
File Edit View Insert Format Tools Add-ons Help
Last edit was made on December 11, 2021 by CANACO JENRY KIM
I UA
E : : = E : E - E - E E
2 Editing
100% -
Normal text
Arial
B
(c) if a constant force just enough to start the crate moving is applied to
it, what would be its acceleration?
- Problem No.2
7. An elevator weighs 1960 N and is being supported by a cable having
a
tension equal to 2300 N. What is the acceleration of the elevator?
9:50 PM
O Type here to search
2 25°C Mostly clear
1/2/2022

f Facebook
x B Physics November 10 Act - Goog x
A MODULE 5 PROBLEM SET
x +
A docs.google.com/document/d/1aREpjEqEtccCsqeKD561vXfHQHHFZNZAG9wvNzcnww0/edit?fbclid=lwAR3Hzj3imvprWw7PYONz-ZKpdd-bD-4u59kRIOPCiKjZi. Q
Physics November 10 Act * e O
A Share
File Edit View Insert Format Tools Add-ons Help
Last edit was made on December 11, 2021 by CANACO JENRY KIM
14 + BI U A
E E = = 1E
E E X
100% - Normal text
Arial
= E - IE
2 Editing
Problem No. 3
- Problem No.2
Two blocks are connected by a massless rope as shown below. The
mass of the block on the table is 4.0 kg and the hanging mass is 1.0 kg
The table and the pulley are frictionless.
Newton's second law:
ΣFEma
Uniform accelerated motion (UAM) equations, a.k.a. "kinematics
equations":
vf = v0 -gtay= (vf+v02)tay= vot-12gt2 vf2= v02 -2gay
(a)
Applying Newton's second law to the block on the table:
EF = ma
T= m,a,
Applying Newton's second law to the hanging block: Tension pulls up,
weight is directed downward, and the acceleration is directed downward.
EF = ma
T- m,g = - m,az
m,a, - m,g = - m,az
m,a, + m,a, = m,g
а, - а, - а
(m, + m2)a = mg
a = m2g/(m1 + m2) = (1 x 9.8)/(4.0 + 1.0) = 9.8/5 = 1.96 m/s2
(b) The acceleration of the system is 1.96 m/s2
9:50 PM
O Type here to search
2 25°C Mostly clear
1/2/2022

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