Question

a string is attached to a block that lies on a horizontal table with a rough surface. in three different trials, the string pulls the block along the table at the same constant velocity to the right. in trial 1, the string is pulled horizontally. in trial 2, the string is pulled at an angle θ₀ above the horizontal. in trial 3, the string is pulled at an angle θ₀ below the horizontal. in which trial, if any, is the string tension the greatest? a trial 1 b trial 2 c trial 3 d the string tension is the same in all three trials.

Explanation:

Step1: Analyze forces in Trial 1

Since the block moves at constant - velocity, the net force is zero. Let the mass of the block be $m$, the coefficient of kinetic friction be $\mu_k$, and the tension be $T_1$. The normal force $N_1 = mg$, and the frictional force $f_1=\mu_kN_1=\mu_kmg$. In the horizontal direction, $T_1 = f_1=\mu_kmg$.

Step2: Analyze forces in Trial 2

The tension is $T_2$. Resolve the tension into horizontal and vertical components. The vertical component of the tension is $T_{2y}=T_2\sin\theta_0$ and the horizontal component is $T_{2x}=T_2\cos\theta_0$. The normal force $N_2=mg - T_2\sin\theta_0$. The frictional force $f_2=\mu_kN_2=\mu_k(mg - T_2\sin\theta_0)$. Since the block moves at constant - velocity, $T_2\cos\theta_0=f_2=\mu_k(mg - T_2\sin\theta_0)$. Solving for $T_2$ gives $T_2=\frac{\mu_kmg}{\cos\theta_0+\mu_k\sin\theta_0}$.

Step3: Analyze forces in Trial 3

The tension is $T_3$. The vertical component of the tension is $T_{3y}=T_3\sin\theta_0$ (down - ward) and the horizontal component is $T_{3x}=T_3\cos\theta_0$. The normal force $N_3=mg + T_3\sin\theta_0$. The frictional force $f_3=\mu_kN_3=\mu_k(mg + T_3\sin\theta_0)$. Since the block moves at constant - velocity, $T_3\cos\theta_0=f_3=\mu_k(mg + T_3\sin\theta_0)$. Solving for $T_3$ gives $T_3=\frac{\mu_kmg}{\cos\theta_0-\mu_k\sin\theta_0}$.

Comparing $T_1=\mu_kmg$, $T_2=\frac{\mu_kmg}{\cos\theta_0+\mu_k\sin\theta_0}$, and $T_3=\frac{\mu_kmg}{\cos\theta_0-\mu_k\sin\theta_0}$, we know that $\cos\theta_0-\mu_k\sin\theta_0<\cos\theta_0<\cos\theta_0+\mu_k\sin\theta_0$ (for $\mu_k>0$ and $0 <\theta_0<90^{\circ}$). So, $T_3>T_1>T_2$.

Answer:

C. Trial 3