Block A of Figure 5 has a mass of 4.00 kg, and block B, 12.0 kg of. The coeffici
ID: 1438082 • Letter: B
Question
Block A of Figure 5 has a mass of 4.00 kg, and block B, 12.0 kg of. The coefficient of kinetic friction between block B and the horizontal surface is 0.25. a) That mass has the block C if B moves to the right with acceleration of 2.00 m/s^2? b)the tension in each string in such a situation? El bloque A de la figura 5 tiene masa de 4.00 kg. y el bloque B, de 12.0 kg. El coeficiente de friccion cinetica entre el bloque B y la superficie horizontal es de 0.25. a) Que masa tiene el bloque C si B se mueve a la derecha con aceleracion de 2.00 m/s^2? b) Que tension hay en cada cuerda en tal situacion?Explanation / Answer
Ans :- First make variables for what you know and don't know:
mA = mass of block A (4kg)
mB = mass of block B (12kg)
mC = mass of block C (= ??)
a = The acceleration of Block B (2 meters/s²)
= coefficient of friction (0.25)
T_ab = Tension in the cord between A and B (=??)
T_bc = Tension in the cord between B and C (=??)
First notice that, since all the blocks are tied together, they are ALL accelerating at the same rate (a= 2 meters/s²). That's important to know.
Apply Newton’s 2nd law all the mass
Fnet = ma
Block A
Tab – mA*g = mA*a.....................1
Block B
Tbc – Tab – mB*g*µ= mB*a................2
BlockC
mC*g – Tbc = mC*a.............................3
From 1
Tab – 4*9.8 = 4*2
Tab = 47.2N
From 2
Tbc - 47.2 – 12*9.8*0.25 = 12*2
Tbc = 100.6N
mC*9.8 – 100.6 = mC*2
mC = 12.90 = 13Kg