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Dima020 [189]
3 years ago
5

Consider the following system, where F = 80 N, m = 1 kg, and M = 11 kg M m F What is the magnitude of the force with which one b

lock acts on the other?
Physics
1 answer:
Lorico [155]3 years ago
8 0

Answer:

Force exerted by the lighter block on the heavier block is 6.63 N

Explanation:

Given Data

F = 80N

m = 1kg

M = 11kg

Solution:

*We assume that there is no friction

Calculating the acceleration of the system

a = \frac{F}{m+M}

a = \frac{80}{1+11}

a = \frac{80}{12}

a = 6.67ms^{-2}

Let's write the Equation of Motion of the heavier block  

F_{1} = F - F_{2}

Ma = F - F_{2}

force exerted by the lighter block on the heavier block is calculated as

 F_{2} = F - Ma

 F_{2} = 80 - (11 x 6.67)  

 F_{2} = 6.63 N

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