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. part b what is the mass of block b? express your answer with the appropriate units. m = value units

Explanation:

Step1: Find acceleration using kinematic equation

The kinematic equation for displacement is $x = v_0t + \frac{1}{2}at^2$. Since the blocks start from rest, $v_0 = 0$. So $x = \frac{1}{2}at^2$. We can solve for $a$:
$$a = \frac{2x}{t^2}$$
Substituting $x = 18.0\ m$ and $t = 5.00\ s$:
$$a = \frac{2\times18.0}{(5.00)^2} = \frac{36.0}{25.0} = 1.44\ m/s^2$$

Step2: Apply Newton's second law to the system

The total mass of the system is $M = m_A + m_B$, where $m_A = 17.0\ kg$ and $m_B$ is the mass we need to find. The net force on the system is $F = 60.0\ N$. By Newton's second law, $F = Ma$. So:
$$M = \frac{F}{a}$$
Substituting $F = 60.0\ N$ and $a = 1.44\ m/s^2$:
$$M = \frac{60.0}{1.44} \approx 41.67\ kg$$

Step3: Solve for mass of block B

Since $M = m_A + m_B$, we can solve for $m_B$:
$$m_B = M - m_A$$
Substituting $M \approx 41.67\ kg$ and $m_A = 17.0\ kg$:
$$m_B = 41.67 - 17.0 = 24.67\ kg$$ (rounded appropriately)

Answer:

The mass of block B is approximately $\boxed{24.7\ kg}$ (or $24.67\ kg$ depending on rounding)