Question 228: Laws of Motion

Question

Three blocks A, B and C, of masses 4 kg, 2 kg and 1 kg respectively, are in contact on a frictionless surface, as shown. If a force of 14 N is applied on the 4 kg block, then the contact force between A and B is

Accepted Answer

Here, M
A
= 4 kg, M
B
= 2 kg, M
C
= 1 kg, F = 14 N
Net mass, M = M
A
+ M
B
+ M
C
= 4 + 2 + 1 = 7 kg
Let a be the acceleration of the system.
Using NewtonÃ¢ÂÂs second law of motion,
F = Ma
14 = 7a    $$ 	herefore $$ a = 2 ms
Ã¢ÂÂ2
Let F' be the force applied on block A by block B
i.e. the contact force between A and B. Free body
diagram for block A
Again using NewtonÃ¢ÂÂs second law of motion,
F Ã¢ÂÂ F' = 4a
14 Ã¢ÂÂ F' = 4 ÃÂ 2 $$ \Rightarrow $$ 14 Ã¢ÂÂ 8 = F' $$ 	herefore $$ F' = 6 N

Answer

8 N

Answer

18 N

Answer

2 N

Answer

6 N

Laws of Motion

Three blocks A, B and C, of masses 4 kg, 2 kg and 1 kg respectively, are in contact on a frictionless surface, as shown. If a force of 14 N is applied on the 4 kg block, then the contact force between A and B is

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