Question

two blocks connected by a light horizontal rope sit at rest on a horizontal, frictionless surface. block a has mass 17.0 kg, and block b has mass m. a constant horizontal force f = 60.0 n is applied to block a (figure 1). in the first 5.00 s after the force is applied, block a moves 18.0 m to the right. while the blocks are moving, what is the tension t in the rope that connects the two blocks? express your answer with the appropriate units.

Explanation:

Step1: Find acceleration using kinematics

We use the kinematic equation $x = v_0t+\frac{1}{2}at^2$. Since initial velocity $v_0 = 0$, the equation simplifies to $x=\frac{1}{2}at^2$. We solve for $a$:
$$a=\frac{2x}{t^2}$$
Substituting $x = 18.0\,\text{m}$ and $t = 5.00\,\text{s}$:
$$a=\frac{2\times18.0}{5.00^2}=\frac{36.0}{25.0}=1.44\,\text{m/s}^2$$

Step2: Find total mass of the system

Let $m_A = 17.0\,\text{kg}$ (mass of block A) and $m_B$ be mass of block B (wait, actually, wait—wait, the problem says "block A has mass 17.0 kg, and block B has mass m"—wait, but maybe we can first find the total mass? Wait, no, first, for the system (A + B), the net force is $F=(m_A + m_B)a$? Wait, no, wait—the force $F = 60.0\,\text{N}$ is applied to block A. Wait, actually, let's correct: the two blocks are connected, so the system is A and B. Let $m_A = 17.0\,\text{kg}$, $m_B = m$ (unknown, but wait, maybe we can find $m$ first? Wait, no—wait, the kinematic data: $x = 18.0\,\text{m}$, $t = 5.00\,\text{s}$, $v_0 = 0$. So acceleration $a = 1.44\,\text{m/s}^2$ as above. Then, for the entire system (A and B), the net force is $F=(m_A + m_B)a$. Wait, but $F = 60.0\,\text{N}$, so:
$$60.0=(17.0 + m_B)\times1.44$$
Solving for $m_B$:
$$17.0 + m_B=\frac{60.0}{1.44}\approx41.6667$$
$$m_B\approx41.6667 - 17.0 = 24.6667\,\text{kg}$$

Step3: Find tension using Newton's second law on block B

The tension $T$ is the net force on block B, so $T = m_Ba$. Substituting $m_B\approx24.6667\,\text{kg}$ and $a = 1.44\,\text{m/s}^2$:
$$T = 24.6667\times1.44\approx35.52\,\text{N}$$

Wait, alternatively, another approach: since the acceleration is the same for both blocks, for block A: $F - T = m_Aa$, and for block B: $T = m_Ba$. Also, from kinematics, $a=\frac{2x}{t^2}=1.44\,\text{m/s}^2$. Then, from $F=(m_A + m_B)a$, we can write $m_B=\frac{F}{a}-m_A$. Then $T = m_Ba = F - m_Aa$. Let's use this formula:
$$T = F - m_Aa$$
Substituting $F = 60.0\,\text{N}$, $m_A = 17.0\,\text{kg}$, $a = 1.44\,\text{m/s}^2$:
$$T = 60.0 - (17.0\times1.44)$$
$$T = 60.0 - 24.48 = 35.52\,\text{N}$$

Answer:

The tension \( T \) is \(\boxed{35.5}\) N (or more precisely \( 35.52 \) N, depending on rounding).