Newton's Second Law Problems With Solutions 2022

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 =
Considering the surfaces and pulley = Frictionless
3