Question

18. (M = 0.5, 4, = 0.4) -
F
7 kg
35 kg
M=
0 -
If a force F of 100 N is applied to block of mass
7 kg, then acceleration of 35 kg mass will be
(1) 2.4 ms-2
(2) 1 ms-2
(3) 0.4 ms-2
(4) 0.8 ms-2​

Answer 1

Answer:

0.8 m/s^2

Explanation:

Given  If a force F of 100 N is applied to block of mass

7 kg, then acceleration of 35 kg mass will be

Force is greater than static friction. So f = 0.5 x 7 x 10 = 35 N

When force is applied the 7 kg block will go in motion and hence there will be kinetic friction. So f = 0.4 x 7 x 10 = 28 N

So taking 35 kg block we have μ = 0

35 a = 28

 a = 28/35

  a = 0.8 m/s^2

acceleration of 35 kg mass will be 0.8 m/s^2

Answer 2

Answer:

Explanation:

i hope... Its clear