A 13 kg block(mA) rests on a level table. The coefficient of kinetic friction between the block and the table is .15. A string of negligible mass is attached to the block. The string, which is perfectly horizontal, passes through a frictionless pulley at the edge of the table and is attached to a suspended 5.0 kg block(mB)

a. Determine the acceleration of each block
b. If mA is initially at rest 1.25 m from the edge of the table, how long does it take to reach the edge of the table if the system is allowed to move freely?
c. If, instead, mB had a mass of 1.0 kg how much mass must mA contain if the acceleration of the system is kept to 1/100 g?

pulling force=5g

retarding force=13g*.15
total mass=18kg

Net force=totalmass*a
5g-13g*.15=(18)a solve for a

b. d=1/2 a t^2 solve for time t.

c. 1*g-mA*g=(1+mA)*g/100
100(1-mA)=1+mA
101mA=99
ma=99/101 check my math.

To solve this problem, we can use Newton's second law (F = ma) and the concept of friction.

a. In this system, there are two forces acting on the 13 kg block (mA): the force due to gravity (weight) and the force of friction. The force due to gravity can be calculated by multiplying the mass (13 kg) by the acceleration due to gravity (9.8 m/s^2). Therefore, the weight of the block is 127.4 N (13 kg × 9.8 m/s^2).

The force of friction can be found by multiplying the coefficient of kinetic friction (0.15) by the normal force. The normal force is equal to the weight of the block when it is on a level surface. Therefore, the normal force is also 127.4 N.

The force of friction is then 0.15 multiplied by 127.4 N, resulting in 19.11 N. Since the system is horizontal and the block is being pulled by the string with the same force as the hanging block, the force of friction is equal to the force of tension in the string.

Now, we can calculate the acceleration of the block (mA) using Newton's second law. The net force on the block is the difference between the force of tension and the force of friction. Therefore, the net force is 19.11 N – force of tension.

Let's assume the acceleration of the system is a. For block mA, the equation becomes:

19.11 N – force of tension = 13 kg × a

For block mB (the hanging block), the force of tension is equal to its weight since it is in equilibrium:

force of tension = 5.0 kg × 9.8 m/s^2

By substituting this value into the equation for block mA, we have:

19.11 N – (5.0 kg × 9.8 m/s^2) = 13 kg × a

Simplifying the equation, we can calculate the acceleration of the system.

b. To determine how long it takes for mA to reach the edge of the table if the system is allowed to move freely, we can use the equation of motion:

d = v_0 t + 0.5 a t^2

where d is the distance traveled, v_0 is the initial velocity (0 m/s since mA is initially at rest), t is the time, and a is the acceleration.

We know that d = 1.25 m and a is the acceleration of the system. To find t, we need to solve for it by rearranging the equation:

0.5 a t^2 = 1.25 m

Substituting the value of a we calculated in part a, we can solve for t.

c. If mB has a mass of 1.0 kg and the acceleration of the system is 1/100 g, we can calculate the total mass (m_total) required for mA to maintain this acceleration.

Given that a = 1/100 g = (1/100) × 9.8 m/s^2, we can use the equations and logic explained in part a to solve for m_total. By substituting the known values into the equation, we can solve for m_total.

Keep in mind that the force of friction will still be equal to the force of tension, and the force of tension will be equal to m_total times the acceleration.

I hope this explanation helps you solve the problem! If you need further clarification or assistance, please let me know.