Analysis of the Problem:

The provided image appears to show a friction experiment using a block with a mass of $1008 \, \text{g}$ (or $1.008 \, \text{kg}$) sliding on a surface of rubber on ice. The graph at the bottom depicts the force measured in Newtons (N) over time while the block is being pulled.

From the graph, the force plateaus around $1.5 \, \text{N}$ during a constant-speed interval (approximately between $t = 4 \, \text{s}$ and $t = 7 \, \text{s}$).

Q3: Net Force Acting on the Block

If the block is being pulled at a constant speed, the acceleration of the block is zero ($a = 0$). According to Newton's second law of motion:

$F_{\text{net}} = m \cdot a$

Since $a = 0$, the net force is: $F_{\text{net}} = 0 \, \text{N}$

Thus, the magnitude of the net force acting on the block is $0 \, \text{N}$.

Q4: Magnitude of the Frictional Force

When the block is moving at a constant speed, the pulling force balances the frictional force ($F_{\text{friction}}$). From the graph, during the constant-speed interval ($t = 4 \, \text{s}$ to $t = 7 \, \text{s}$), the force plateaus at approximately:

$F_{\text{pull}} = 1.5 \, \text{N}$

Since the block is in equilibrium (no acceleration), the pulling force is equal to the frictional force: $F_{\text{friction}} = F_{\text{pull}} = 1.5 \, \text{N}$

Thus, the magnitude of the frictional force is $1.5 \, \text{N}$.

Summary of Answers:

1. Net Force Acting on the Block: $0 \, \text{N}$
2. Frictional Force Resisting the Motion: $1.5 \, \text{N}$

Would you like further details or help understanding these results?

Related Questions:

1. What is the coefficient of friction between the surfaces, given the normal force?
2. How does the mass of the block affect the frictional force?
3. What would happen to the net force if the block were accelerating instead of moving at a constant speed?
4. How can we calculate the normal force acting on the block in this scenario?
5. Why does the frictional force plateau during the constant-speed phase?

Tip:

The frictional force acting on an object is always proportional to the normal force, given by $F_{\text{friction}} = \mu \cdot F_{\text{normal}}$, where $\mu$ is the coefficient of friction.