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The coefficient of static and kinetic frictions between a 15.2-kg box and a desk are 0.55 and 0.36, respectively. What is the net force on the box when a 120.9 N horizontal force is applied to the box while moving?

1 Answer

6 votes

Answer:

67.28 N

Step-by-step explanation:

Given that,

The mass of an box, m = 15.2 kg

The coefficients of static and kinetic frictions for plastic on wood are 0.55 and 0.36, respectively.

The force of static friction,


F_s=\mu_smg\\\\F_s=0.55* 15.2* 9.8\\\\F_s=81.92\ N

The force of kinetic friction,


F_k=\mu_kmg\\\\F_k=0.36* 15.2* 9.8\\\\F_k=53.62\ N

Net force acting on the object is :

F = 120.9 -53.62

= 67.28 N

Hence, this is the required solution.

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