Question

87. A 4-kg block and a 2-kg block can move on the
horizontal frictionless surface. The blocks are
accelerated by a +12-N force that pushes the larger
block against the smaller one. Determine the force that
the 2-kg block exerts on the 4-kg block.
(a) -4N
(C) ON
(b) -12 N
(d) +4 N
(e) +8 N​

Answer 1

Answer:

Answer:

F_n = -4 NF

n

=−4N

Explanation:

Net external force that exerted on the block is given as

F_{net} = (m_1 + m_2) aF

net

=(m

1

+m

2

)a

here we know that

F = 12 NF=12N

m_1 = 4 kgm

1

=4kg

m_2 = 2 kgm

2

=2kg

now we have

12 = (4 + 2) a12=(4+2)a

so we have

a = 2 m/s^2a=2m/s

2

now the force exerted by bigger block on smaller block is given as

F_n = maF

n

=ma

F_n = (2 \times 2)F

n

=(2×2)