Question

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) 4 n (b) 8 n (c) 20 n (d) 40 n (e) 80 n 6. 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 a hanging block

For each of the 4 - kg hanging blocks, using Newton's second - law \(F = ma\). In the vertical direction, since the blocks are not moving (\(a = 0\)), the tension in the vertical part of the rope \(T\) balances the weight of the block. The weight of a 4 - kg block is \(W=mg\), where \(m = 4\space kg\) and \(g = 10\space m/s^{2}\), so \(W=4\times10 = 40\space N\), and \(T = 40\space N\).

Step2: Analyze horizontal part of the rope

The horizontal part of the rope is in equilibrium. Consider the forces acting on the horizontal part of the rope. The two vertical tensions pulling on the horizontal part of the rope are equal in magnitude and opposite in the horizontal - component direction. Since the system is in equilibrium (the blocks are not moving), the tension in the horizontal part of the rope must also be \(T_{h}=40\space N\).

for the second question:

Step1: Find the acceleration of the whole system

The total mass of the system of three blocks is \(m_{total}=m + 3m+2m=6m\). According to Newton's second - law \(F = m_{total}a\), the acceleration of the whole system when the force \(F\) is applied is \(a=\frac{F}{6m}\).

Step2: Analyze the force on each spring

For spring \(A\), the force applied to it is \(F\). For spring \(B\), the force it exerts is to accelerate the mass \(3m + 2m=5m\). Using \(F = ma\), the force on spring \(B\) is \(F_{B}=(3m + 2m)a=(3m + 2m)\times\frac{F}{6m}=\frac{5}{6}F\). For spring \(C\), the force it exerts is to accelerate the mass \(2m\), so \(F_{C}=2m\times a=2m\times\frac{F}{6m}=\frac{1}{3}F\).

Step3: Use Hooke's law

According to Hooke's law \(F = kx\) (where \(k\) is the spring - constant and \(x\) is the extension). Since the springs are identical (\(k\) is the same for all), the spring with the largest force applied to it will stretch the most. Since \(F>F_{B}>F_{C}\), spring \(A\) is stretched the most.

Answer:

D. 40 N