Physics Midterm #2 (Spring 2020): Newton's Laws Problem Solutions

This document presents the solutions to a Physics Midterm #2 examination, focusing on problems related to Newton's Laws of motion. The solutions encompass various scenarios, including frictionless surfaces, tension in strings, and the application of Newton's second law to calculate forces and accelerations in multi-body systems. Detailed step-by-step solutions are provided for each problem, incorporating free body diagrams and relevant equations. The problems cover topics such as calculating tension in strings, determining acceleration of objects, analyzing forces on inclined planes, and considering friction. Each solution includes the application of Newton's laws, force calculations, and the derivation of final answers, providing a comprehensive guide to solving these types of physics problems. This resource is designed to aid students in understanding and mastering the concepts of Newtonian mechanics.

PHYSICS
Spring 2020-Midterm#2

Problem 1
Horizontal surface is given as frictionless.
F = 12 N
Tension in string 1 and string 2 =?
Assuming that M, 2M and 3Mall the three bodies as a whole one body and hence, the net
force on the system would be F in rightward direction as given below.
Net force=3 F−2 F=F
Net acceleration= F
M +2 M +3 M = F
6 M
Let, tension in string 1 be T1 and in string 2 be T2.
Now,
Applying Newton’s second law on block 3M
3 F−T 1=3 M∗F
6 M
T 1=3 F− F
2
T 1=2.5 F=2.5∗12=30 N
Further,
Applying Newton’s second law on block M
T 2−2 F= M∗F
6 M
T 2=2 F + F
6 = 13
6 F=13
6 ∗12=26 N
1

Therefore, the tension in string 1 would be 30 N and in string 2 would be 26 N.
Problem 2
Mass of m1 = 2kg
Mass of m2 = 6kg
Angle = 55 degree
(a) Freebody diagram
(b) Magnitude of acceleration of objects
Applying Newton’s second law
2

−M 1 g+T =M 1a
−M 2 g sin θ−T =M 2 a
Putting the equations together
−M 1 g+T + ( −M 2 g sin θ−T ) =( M 1+ M 2) a
(−2∗9.8 )− ( 6∗9.8∗sin 55 )= ( 2+6 ) a
a=3.57 m
s2
(c) Tension in string
−M 1 g+T =M 1a
T =M 1 a+ M 1 g
T = ( 2∗3.57 ) + ( 2∗9.8 )=26.7 N
(d) Speed of objects 2 second after their released from rest position
As the block released from rest then the initial velocity of the respective block would be zero.
u=0 m
s
Further, the final velocity of blocks would be computed through the equation of motion as
shown below.
v=u+at
v=0+ ( 3.57∗2 )=7.14 m
s
Problem 3
Mass of block M = 2.2 kg
Tension in connecting string =
3

Considering the surfaces and pulley = Frictionless
Here,
Mass of block over inclination
Mg sin θ+T =Ma
Hanging mass of block
Mg−T =Ma
Solving the two equation
Mg sin θ+T =Mg−T
T = Mg−Mg sin θ
2 =2.2∗9.8 ¿ ¿
T =5.39∨5.4 N
Tension in connecting string would be 5.4 N.
Problem 4
Mass of block = 3 kg
Angle with wall = 50 degree
Coefficient of static friction between wall and block = 0.250
4

(a) Possible values of magnitude of P to keep the block static
Possible values of magnitude of P to keep the block static would be 31.72 N and 48.57 N.
Problem 5
Coefficient of friction between incline and block μk = 0.40
Magnitude of acceleration of suspended block =?
5

Freebody diagram
As shown in the FBD the mass 2M is acting through tension T in upward direction and the
gravity force is acting in downward direction.
Applying Newton’s second law
2 Mg−T =2 Ma
Or
T =2 M ( g−a )
Further, normal force N is acting on mass M and hence,
N=Mg cos θ
Where,
θ=Angle of inclination
It is given that M is acting through kinetic friction and therefore,
Force of kinetic friction
6

Fk=μk N
Fk=μk .( Mg cos θ)
After considering the force equation on M
F=ma
Also, M is acting through tension, force of gravity and friction force and therefore,
Applying Newton’s second law
−Mg sin θ−Fk+T =Ma
Putting the value of Fk∧T
−Mg sin θ−μk . ( Mg cos θ ) +2 M ( g−a ) =Ma
a=−g sinθ−μk . ( g cos θ ) +2 g
3
Also putting θ=40 °
μk =0.40
g=9.8
Hence,
a=−9.8 sin 40−0.40 ( 9.8 cos 40 ) + ( 2∗9.8 )
3
a=3.43 m
s2
Magnitude of acceleration of suspended block is 3.43 m per s^2.
Problem 6
The weight of m1 = 4kg
The weight of m2 = 1kg
The weight of m3 = 2kg
7

Freebody diagram
Let a is positive magnitude of the acceleration and -aj in m1, -ai in m2 and +aj in m3. Let the
tension in left side cord is T12 and tension in right side cord is T23.
The force equations can be drawn based on above
8

(a) Acceleration in each object along with direction
After adding the above three equations
39.2 N −3.43 N−19.6 N =7 kg∗( a )
a=2.31 m
s2
Hence,
Magnitude of the acceleration in m1 = 2.31 m/s^2 (In downward direction)
Magnitude of the acceleration in m2 = 2.31 m/s^2 (In left side)
Magnitude of the acceleration in m3 = 2.31 m/s^2 (In upward direction)
(b) Tension in two cords
From above equation
−T 1 2+39.2 N =4 a
−T 12+39.2 N=4∗2.31
𝑇12=30 𝑁
Further,
T 23−19.6 N=2 a
T 23−19.6 N=2∗2.31
T 23=24.2 N
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