The diagram below shows block A, of mass x, attached to block B, of mass 2x, by a light, inextensible string that passes over a frictionless pulley. A constant force of 40 N, applied at an angle of 30° to the horizontal, accelerates block B to the right at 1.5 m s².

The vertical forces are in equilibrium, so the sum of upward forces equals the sum of downward forces.

$$\sum F_y = 0$$ $$N + F_{app} \sin \theta - W_B = 0$$

Here, $W_B = m_B g = 2xg$, $F_{app} = 40 \, \text{N}$, $\theta = 30^\circ$, and $g = 9.8 \, \text{m/s}^2$.

$$N + 40 \sin 30^\circ - 2x(9.8) = 0$$ $$N + 40(0.5) - 19.6x = 0$$ $$N + 20 - 19.6x = 0$$ $$N = 19.6x - 20 \, \text{N}$$

The formula for kinetic friction is $f_k = \mu_k N$.
Given $\mu_k = 0.2$.

$$f_k = 0.2 (19.6x - 20)$$ $$f_k = 0.2 \times 19.6x - 0.2 \times 20$$ $$f_k = 3.92x - 4 \, \text{N}$$

Thus, the magnitude of the kinetic friction is $\boxed{(3.92x - 4) \, \text{Newtons}}$.

2.3 State Newton's second law of motion in words.
Newton's second law of motion states that the net force acting on an object is equal to the product of its mass and acceleration, and the direction of the acceleration is the same as the direction of the net force.

2.4 Apply Newton's second law of motion to each block separately to determine the mass of block B.

Block A has mass $m_A = x$ and accelerates upwards at $a = 1.5 \, \text{m/s}^2$.
Forces on Block A: Tension $T$ (upwards), Weight $W_A = m_A g = xg$ (downwards).

$$\sum F_A = m_A a$$ $$T - W_A = m_A a$$ $$T - x(9.8) = x(1.5)$$ $$T = 1.5x + 9.8x$$ $$T = 11.3x \, \text{N} \quad \text{(Equation 1)}$$

Block B has mass $m_B = 2x$ and accelerates to the right at $a = 1.5 \, \text{m/s}^2$.
Horizontal forces on Block B: $F_{app,x} = F_{app} \cos \theta$ (right), Tension $T$ (left), Kinetic friction $f_k$ (left).

$$\sum F_B = m_B a$$ $$F_{app} \cos \theta - T - f_k = m_B a$$

Substitute $F_{app} = 40 \, \text{N}$, $\theta = 30^\circ$, $f_k = (3.92x - 4) \, \text{N}$ (from 2.2), and $m_B = 2x$.

$$40 \cos 30^\circ - T - (3.92x - 4) = 2x(1.5)$$ $$40\left(\frac{\sqrt{3}}{2}\right) - T - 3.92x + 4 = 3x$$ $$20\sqrt{3} - T - 3.92x + 4 = 3x$$

$$20\sqrt{3} - (11.3x) - 3.92x + 4 = 3x$$ $$20\sqrt{3} + 4 = 3x + 11.3x + 3.92x$$ $$20\sqrt{3} + 4 = (3 + 11.3 + 3.92)x$$ $$20\sqrt{3} + 4 = 18.22x$$ $$x = \frac{20\sqrt{3} + 4}{18.22}$$ $$x \approx \frac{20(1.73205) + 4}{18.22}$$ $$x \approx \frac{34.641 + 4}{18.22}$$ $$x \approx \frac{38.641}{18.22}$$ $$x \approx 2.1208 \, \text{kg}$$

The mass of Block B is $m_B = 2x$.

$$m_B = 2 \times 2.1208$$ $$m_B \approx 4.2416 \, \text{kg}$$

The mass of block B is approximately $\boxed{4.24 \, \text{kg}}$.

2.5 The angle between the applied force on block B and the horizontal is now decreased. How will this change affect the following?

2.5.1 The coefficient of kinetic friction between block B and the surface.
The coefficient of kinetic friction ($\mu_k$) depends only on the nature of the surfaces in contact, not on the angle of the applied force.

$$\text{REMAINS THE SAME}$$

2.5.2 The normal force acting on block B.
From part 2.2, the normal force is $N = W_B - F_{app} \sin \theta$.
If the angle $\theta$ decreases (assuming $0^\circ < \theta < 90^\circ$), then $\sin \theta$ will decrease.
Since $F_{app} \sin \theta$ is being subtracted from $W_B$, a smaller value of $F_{app} \sin \theta$ will result in a larger normal force $N$.

$$\text{INCREASES}$$