Question

1. two masses are connected together by a rope and pulley on a frictionless incline plane as shown. when the system is released, what is the initial acceleration (magnitude and direction) of the 21 kg mass? 21(9.8)sin65 8.9.8 108.2 / 29 = 3.43 m/s² 7. two packing crates of masses 10.0 kg and 5.00 kg are connected by a light string that passes over a frictionless pulley. the 5.00 - kg crate lies on a smooth incline of angle 40.0°. find the acceleration of the 5.00 - kg crate and the tension in the string. 5(9.8)=fgsin40 10(9.8 =)fg m1 (5kg) m2 = 10kg 65.5 / 15 = 4.43m/s² 8. a 28.0 - kg block is connected to an empty 1.35 - kg bucket by a cord running over a frictionless pulley. the coefficient of friction between the table and the block is 0.320. sand is gradually added to the bucket until the system just begins to move. (a) calculate the mass of sand added to the bucket. (b) calculate the acceleration of the system if 8.00 kg were added to the system.

Explanation:

Step1: Analyze forces for problem 8 part (a)

When the system just begins to move, the tension in the cord equals the frictional - force on the block. The frictional force on the block is $F_f=\mu_kN$, where $N = m_{block}g$ (since the block is on a horizontal surface and the normal force equals the weight of the block). So $F_f=\mu_km_{block}g$. Let the mass of the sand be $m_{sand}$, and the total mass of the bucket and sand is $m = m_{bucket}+m_{sand}$. The tension $T = mg=(m_{bucket}+m_{sand})g$. Setting $F_f = T$, we have $\mu_km_{block}g=(m_{bucket}+m_{sand})g$. Canceling out $g$ on both sides, we get $m_{sand}=\mu_km_{block}-m_{bucket}$.
Given $\mu_k = 0.320$, $m_{block}=28.0$ kg, and $m_{bucket}=1.35$ kg.
$m_{sand}=0.320\times28.0 - 1.35$.
$m_{sand}=8.96 - 1.35=7.61$ kg.

Step2: Analyze forces for problem 8 part (b)

The total mass of the system is $M=m_{block}+m_{bucket}+m_{added - sand}=28.0 + 1.35+8.00=37.35$ kg.
The net - force acting on the system is $F_{net}=(m_{bucket}+m_{added - sand})g-\mu_km_{block}g$.
$F_{net}=(1.35 + 8.00)\times9.8-0.320\times28.0\times9.8$.
$F_{net}=(9.35)\times9.8-0.320\times28.0\times9.8$.
$F_{net}=9.35\times9.8-8.96\times9.8=(9.35 - 8.96)\times9.8=0.39\times9.8 = 3.822$ N.
According to Newton's second law $F_{net}=Ma$, so $a=\frac{F_{net}}{M}$.
$a=\frac{3.822}{37.35}\approx0.102$ m/s².

Answer:

(a) $7.61$ kg
(b) $0.102$ m/s²