Question

two force of 40 N and 72 N are applied separately on a body of mass 8 kg. find the acceleration produced in each case​

Answer 1

Explanation:

Horizontal force,F=600 N

Mass of body A, m

1

=10 kg

Mass of body B, m2=20 kg

Total mass of the system, m=m1+m2=30 kg

Using Newtons second law of motion, the acceleration (a) produced in the system can be calculated as:

F=ma

a=

m

F

=

30

600

=20m/s

2

When force F is applied on body A:

The equation of motion can be written as: F−T=m

1

a

T=F−(m

1

)a

=600−10×20=400 N (i)

When force F is applied on body B:

The equation of motion can be written as:

F−T=m

2

a

T=F−(m

2

)a

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.

solution

Answer 2

Answer:

Explanation:

f=ma

a=f/m

1st case = 40/8=5N/Kg^2

2nd case= 72/8 = 9N/Kg^2