Question

unit 2 exam mcq portion (10 question version)
(a) zero
(b) 6n
(c) 8n
(d) 16n
(e) 24n

1. two 4 kg blocks hang from a rope that passes over two frictionless pulleys, as shown in the figure above. what is the tension in the horizontal portion of the rope if the blocks are not moving and the rope and the two pulleys have negligible mass?

(a) 4n
(b) 8n
(c) 20n
(d) 40n
(e) 80n

1. three blocks of masses m, 3m, and 2m resting on a frictionless horizontal surface are connected to identical ideal springs, as shown above. a force of magnitude f directed to the left is then applied to the left end of spring a. which spring is stretched the most when the blocks are all moving with the same acceleration?

(a) a
(b) b
(c) c
(d) none, because the springs do not stretch.
(e) none, because the springs all stretch the same amount.

Explanation:

Step1: Analyze forces on one block

For a single 4 - kg block hanging, the force due to gravity is given by $F = mg$. Here, $m = 4\ kg$ and $g=10\ m/s^{2}$, so $F=4\times10 = 40\ N$. Since the block is in equilibrium (not moving), the tension in the rope must balance the force due to gravity acting on the block.

Step2: Determine tension in horizontal rope

The tension in the rope is the same throughout the mass - less, friction - less rope. So the tension in the horizontal portion of the rope is equal to the tension required to balance the weight of one of the hanging blocks. So the tension $T = 40\ N$.

For the second problem:

Step1: Find the acceleration of the system

The total mass of the system is $m + 3m+2m=6m$. According to Newton's second law $F = ma$, the acceleration of the system $a=\frac{F}{6m}$.

Step2: Analyze the force on each spring

For spring A, the force causing its stretch is $F$. For spring B, the force causing its stretch is the force required to accelerate the mass $3m + 2m=5m$. Using $F = ma$, the force on spring B is $F_B=(3m + 2m)a=(5m)\times\frac{F}{6m}=\frac{5F}{6}$. For spring C, the force causing its stretch is the force required to accelerate the mass $2m$. Using $F = ma$, the force on spring C is $F_C = 2m\times\frac{F}{6m}=\frac{F}{3}$. Since $F>\frac{5F}{6}>\frac{F}{3}$, spring A is stretched the most.

Answer:

1. D. 40 N
2. A. A