Answer to Question #296670 in Molecular Physics | Thermodynamics for dandan

Question

 Two bodies A and B in figure are separated by a spring. Their motion down the incline is resisted by a force P = 800 N . The coefficient of kinetic friction is 0.30 under A and 0.10 under under B. a.) compute the acceleration of block A b.) Compute the acceleration of block B c.) Determine the force in the spring. *

Answer 1

Solution;

Take the mass of A=400N and B=600N

Apply equilibrium for vertical forces;

$\sum F _y=0$

For A;

$N_A-400cos\theta =0$

$N_A-400×\frac 45=0$

$N_A=320N$

For B;

$N_B-600cos\theta=0$

$N_B-600×\frac45=0$

$N_B=480N$

Frictional force is given as;

$f_k=\mu N$

For A;

$f_{kA}=0.30×320=96N$

For B;

$f_{kB}=0.10×480=48N$

Also;

$W=mg$

For A;

$m_A=\frac Wg =\frac{400}{9.81}=40.77kg$

For B;

$m_B=\frac{W_B}{g}=\frac{600}{9.81}=61.16kg$

Taking the system as whole;

$1000sin\theta-200-(f_{kA}+f_{kB})=(m_A+m_B)a$

Substitute the values;

$(1000×\frac 35)-200-(96+48)=(40.77+61.16)a$

$a=3.98m/s^2$

Considering mass A;

$400sin\theta -F_{sp}-96=40.77×3.98$

$F_{sp}=41.74N$

Answers;

a)3.98m/s²

b)3.98m/s²

c)=41.74N