a 9.50kg hanging object.. PHYSICS HELP

Question

A 9.50-kg hanging object is connected by a light, inextensible cord over a light, frictionless pulley to a 5.00-kg block that is sliding on a flat table. Taking the coefficient of kinetic friction as 0.187, find the tension in the string. (The block slides to the right in the diagram below.)

Answer 1

From newtons 2nd law of motion, F_net=ma. Since you have two objects (table weight we'll call object 1 and the hanging weight we'll call object 2) each object needs its own free body diagram and net force, but the mass will be the combined mass (m₁ + m₂) and the acceleration will be the same value.

The equation for the hanging mass is easier to start with since it only has forces in the vertical direction. For the sake of this problem, lets call the direction of motion to be the positive direction. This means the hanging, which is moving down, to have a positive gravitational force (F_g = mg) and a negative tension force acting on it.

(eq1) m₂g - T = (m₁ + m₂)a

For the table weight, since the acceleration in the vertical direction is 0, the normal force and the gravitational force is equal. Thus we can use

(eq2) N₁ = m₁g

In the horizontal direction the table weight has friction acting negatively and tension pulling it forward.

(eq3) T - F_f = (m₁ + m₂)a

Because you are given a coefficient of friction, you can substitute the equation F_f = μ*N₁ and because you already have an equation for N₁ you can substitute equation 2 so that Ff = μ * m₁*g. Putting this all together you can rewrite equation 3 as

(eq4) T - μ*m₁*g = (m₁ + m₂)a

Now with equations 1 and 4 you have two equations with only two unknowns: a and T. From here it can be solved using any system of equation method. Personally, I like to add the two equations together, which cancels out the tension and solve for acceleration first, then substitute to solve for T. This is usually easier and faster though less accurate due to rounding errors. By adding the two equations together you get

(eq5) m₂*g -μ*m₁*g = 2(m₁ + m₂)a

This equation itself shows gravity of the hanging weight pulling the system forward and the friction of the table weight slowing it down.

By substituting your known masses and coefficient of friction you can solve for acceleration: a = 2.89m/s²

Now you can substitute your value of acceleration into either equation 1 or equation 4 to solve for tension. T = 51.1 N