Question

Two blocks connected by a mass less string slide down an inclined plane having angle of inclination ${37}^{\circ }$. The masses of the two blocks are $M_1=4\text{kg}$ and $M_2=2\text{kg}$ respectively and the coefficients of friction 0.75 and 0.25 respectively-

a. The common acceleration of the two masses is $1.33{\text{ms}}^{-2}$
b. The tension in the string is $14.7N$
c. The common acceleration of the two masses is $2.94{\text{ms}}^{-2}$
d. The tension in the string is $5.3N$

Choose the correct statement.

• A. a,d
• B. c,d
• C. b,d
• D. b,c

Answer 1

The correct option is A a,d

$f_1=0.75\times mg\cos {37}^{\circ }$
$f_2=0.25\times mg\cos {37}^{\circ }$
$6g\sin {37}^{\circ }-f_1-f_2=6a$
$6g\sin {37}^{\circ }-0.75\times 4g\cos {37}^{\circ }-0.25\times 2g\cos {37}^{\circ }=6a$
$\frac{6g\sin {37}^{\circ }-0.75\times 4g\cos {37}^{\circ }-0.25\times 2g\cos {37}^{\circ }}{6}=a$
$a=1.33{\text{m/s}}^2$

friction $=0.25\times 2g\cos {37}^{\circ }$
$2g\sin {37}^{\circ }-T-0.25\times 2g\cos {37}^{\circ }=2a$
$T=2g\sin {37}^{\circ }-0.25\times 2g\cos {37}^{\circ }-2\times 1.3$
$T=5.3N$