block with mass m1= 4.30 kg and a ball with mass m2=7.70kg are connected by ligh

Question

block with mass m1= 4.30 kg and a ball with mass m2=7.70kg are connected by light string that passes over a frictionless pulley, as shown in the picture below. T coefficient of kinetic friction between the block and the surface is 0.300. a.) Find the acceleration of the 2 objects and the tension in the string. a= m/s^2 T= N b.) Use the value from the part A to help you work this exercise. What if an additional mass is attached to the ball? How large must this mass be to increase the downward acceleration by 30%?

Explanation / Answer

taking the entire system of M1 + M2 :

the forces acting on the system are :

1) Fk 2) M2g

since M1g is balance by normal reaction force and

Tension is an internal force in the system of M1 + M2 so we need not consider tension for now :

so :

Fk = 0.3*M1g = 0.3*4.3*9.8 = 12.642 N

M2g = 7.7*9.8 = 75.46 N

therefore net force on the system will be M2g - Fk = 62.818 N4

therefore :

62.818 = (M1+ M2)*a

a = 5.2348 m/s^2

now to find the tension in the string we consider any of the masses M1 or M2 :

considering M1 :

the net force on M1 will be M1*a

M1*a = T - Fk

therefore T = M1*a + Fk = 35.1517 N

a = 5.2348

to increase the acceleration by 30% we need to make

a = 5.2348*0.3 + 5.2348 = 6.805 m/s^2

therefore let the new mass added be M3

so the net force on the system of M1, M2 and M3 will be :

(M1+M2+M3)a = (M2+M3)g - Fk

we know M1, M2 ,a , Fk so we find M3

which comes out as:

M3 = 6.28 Kg

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