Question

1. the man pulls on a rope attached to a 30.0 kg mass. if he exerts a force of 500.0 n, determine the acceleration of the system and the tension in the rope connecting the two masses.

Explanation:

Step1: Analyze forces on 30 - kg mass

The net - force on the 30 - kg mass is $F - T_2-\mu N$, where $F = 500N$, $\mu=0.25$, $N = m_1g=30\times9.8 = 294N$. So the net - force is $F - T_2-\mu m_1g$. According to Newton's second law $F - T_2-\mu m_1g=m_1a$.

Step2: Analyze forces on 20 - kg mass

The net - force on the 20 - kg mass is $T_2 - m_2g$. According to Newton's second law $T_2 - m_2g=m_2a$.

Step3: Express $T_2$ from the second equation

$T_2=m_2(g + a)=20(9.8 + a)$.

Step4: Substitute $T_2$ into the first equation

$500-20(9.8 + a)-0.25\times30\times9.8=30a$.
Expand the left - hand side: $500-(196 + 20a)-73.5 = 30a$.
$500-196 - 73.5-20a=30a$.
$230.5=50a$.

Step5: Solve for acceleration $a$

$a=\frac{230.5}{50}=4.61m/s^2$.

Step6: Solve for tension $T_2$

Substitute $a = 4.61m/s^2$ into $T_2=m_2(g + a)$.
$T_2=20\times(9.8 + 4.61)=20\times14.41 = 288.2N$.

Answer:

The acceleration of the system is $4.61m/s^2$ and the tension in the rope is $288.2N$.