plzz solve this ques...
Attachments:
Answers
Answered by
0
Answer:-
Horizontal force, F = 600 N
Mass of body A, m1 = 10 kg
Mass of body B, m2 = 20 kg
Total mass of the system, m = m1 + m2 = 30 kg
Using Newton’s second law of motion, the acceleration (a) produced in the system can be calculated as:
F = ma
∴ a = F / m = 600 / 30 = 20 ms-2
When force F is applied on body A:
The equation of motion can be written as:
F – T = m1a
∴T = F – m1a
= 600 – 10 × 20 = 400 N … (i)
When force F is applied on body B:
The equation of motion can be written as:
F – T = m2a
T = F – m2a
∴T = 600 – 20 × 20 = 200 N … (ii)
which is different from value of T in case (i). Hence our answer depends on which mass end, the force is applied.
Horizontal force, F = 600 N
Mass of body A, m1 = 10 kg
Mass of body B, m2 = 20 kg
Total mass of the system, m = m1 + m2 = 30 kg
Using Newton’s second law of motion, the acceleration (a) produced in the system can be calculated as:
F = ma
∴ a = F / m = 600 / 30 = 20 ms-2
When force F is applied on body A:
The equation of motion can be written as:
F – T = m1a
∴T = F – m1a
= 600 – 10 × 20 = 400 N … (i)
When force F is applied on body B:
The equation of motion can be written as:
F – T = m2a
T = F – m2a
∴T = 600 – 20 × 20 = 200 N … (ii)
which is different from value of T in case (i). Hence our answer depends on which mass end, the force is applied.
Similar questions