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The coefficient of static friction between m1 and the horizontal surface is 0.50, and the coefficient of kinetic friction is 0.30. (a) If the system is released from rest, what will its acceleration be

User PGOnTheGo
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Final answer:

The acceleration of the system can be determined using Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the net force is given by the force of tension in the system, which is equal to the force of gravity acting on m1 minus the force of kinetic friction between m1 and the surface.

Step-by-step explanation:

The acceleration of the system can be determined using Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the net force is given by the force of tension in the system, which is equal to the force of gravity acting on m1 minus the force of kinetic friction between m1 and the surface. The force of kinetic friction can be calculated by multiplying the coefficient of kinetic friction by the normal force, which is equal to the weight of m1. Therefore, the equation for the net force is:

Fnet = m₁a = m₁g - μk(m₁g)

Simplifying this equation, we get:

a = g - μkg

Substituting the given values, where μk is the coefficient of kinetic friction, and g is the acceleration due to gravity:

a = (9.8 m/s²) - (0.30)(9.8 m/s²) = 6.86 m/s²

Therefore, the acceleration of the system will be 6.86 m/s².

User Lotus
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