QUESTION 4 Three identical blocks of mass m on a frictionless surface are connected by massless taut strings. A force Fis applied on the rightmost block as shown in the figure below so that each block moves to right with acceleration a. F T2 m T1 m m Applying Newton's second law on the rightmost block: EFx = = ma + Applying Newton's second law on the block at the center: EFx = - T2 = Applying Newton's second law on the leftmost block: EFx = = ma If each block has a mass of 10 kg and is accelerating at 2 m/s2, the magnitude of the tension on the second string is T2 = N.

Elements Of Electromagnetics
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ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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QUESTION 4
Three identical blocks of mass m on a frictionless surface are connected by massless taut strings. A force Fis
applied on the rightmost block as shown in the figure below so that each block moves to right with acceleration a.
T2
T1
m
F
Applying Newton's second law on the rightmost block:
EFx =
= ma
Applying Newton's second law on the block at the center:
EFx =
|- T2 =
Applying Newton's second law on the leftmost block:
EFx =
= ma
If each block has a mass of 10 kg and is accelerating at 2 m/s-, the magnitude of the tension on the second string is
T2=
N.
Transcribed Image Text:QUESTION 4 Three identical blocks of mass m on a frictionless surface are connected by massless taut strings. A force Fis applied on the rightmost block as shown in the figure below so that each block moves to right with acceleration a. T2 T1 m F Applying Newton's second law on the rightmost block: EFx = = ma Applying Newton's second law on the block at the center: EFx = |- T2 = Applying Newton's second law on the leftmost block: EFx = = ma If each block has a mass of 10 kg and is accelerating at 2 m/s-, the magnitude of the tension on the second string is T2= N.
QUESTION 5
= 1000 kg) at v1 = 11.11 m/s when he stepped on the brakes. The car traveled a distance
Procopio is driving a car (m
of s = 15 m before stopping completely. How much force F was needed to stop the car?
The work done by the force F as the car traveled a distance s before stopping is:
W =
By the work-energy theorem:
2
W =
- K1 = -½
Combining these two equations for work and isolating F, we obtain an expression for F:
F =
2/0
Plugging in values, we know that
4.4 newtons of force is needed to stop the car.
Transcribed Image Text:QUESTION 5 = 1000 kg) at v1 = 11.11 m/s when he stepped on the brakes. The car traveled a distance Procopio is driving a car (m of s = 15 m before stopping completely. How much force F was needed to stop the car? The work done by the force F as the car traveled a distance s before stopping is: W = By the work-energy theorem: 2 W = - K1 = -½ Combining these two equations for work and isolating F, we obtain an expression for F: F = 2/0 Plugging in values, we know that 4.4 newtons of force is needed to stop the car.
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