Question

4. The figure(Figure 1) shows a block (mass mi) on a rough horizontal surface, connected by a thin cord that passes over a pulley to a second block (my), which hangs vertically If ma is 30Kg and mois 40kg, in case the kinetic friction Coefficient is 0.25 what is the acceleration. Use Newton's 2nd Law to find the formula for acceleration first. thanks MA MB A. Write the appropriate Formulas B. Plug in the given data with their units C. Solve for the unknown D. Box the Final Answer with it's unit

Answer 1

Given mA = 30kg, mB = 40kg and coefficient of kinetic friction $\mu$ = 0.25.

(A) Let T be the tension in the string, f be the frictional force acting on the block of mass mA and a be the common acceleration of the two blocks.

The second law equation for the mass mA is,

T =m4a

$T-\mu m_{A}g=m_{A}a$

The second law equation for the mass mB is,

$m_{B}g-T=m_{B}a$

(B) Substituting the given data in the equation for mA, we get,

$T-0.25\times 30kg\times 9.8m/s^{2}=30kg\times a$

$T-73.5N=30kg \times a$

Substituting the given data in the equation for mB, we get,

$40kg \times 9.8m/s^{2}-T=40kg\times a$

$392N-T=40kg\times a$

(C) Adding both the equations, we get

$392N-73.5N=(40kg+30kg)a$

$318.5N=70kg\times a$

$a=\frac{318.5N}{70kg}=4.55m/s^{2}$

$T=30kg \times 4.55m/s^{2}+73.5N$

$T=136.5N+73.5N=210N$

(D) So the acceleration of the system is a = 4.55m/s2 and the tension in the string is T = 210N.