Question

laboratory
conceptual activity
quantitative activity
inclined plane
acceleration = 1.5 m/s²
net force = 1.5 n
mass of block
angle of inclination
coefficient of kinetic friction
coefficient of static friction
exploration 2: net force v. angle of incline
procedure

1. set mass of block to 1.00 kg, coefficient of kinetic friction to 0.40, and coefficient of static friction to 0.50. record

the mass in table 2.

1. with these settings, observe and record the net force for at least five different angles of incline in table 2.
2. change mass of block to 2.00 kg and repeat step 2.

observations and analysis
table 2 (μs = 0.50; μk = 0.40)

Explanation:

To solve for the net force on a block on an inclined plane, we use the formula for net force considering gravity, normal force, and friction.

Step 1: Identify Forces

The forces acting on the block are:

• Gravitational force component down the incline: \( F_{g\parallel} = mg\sin\theta \)
• Frictional force up the incline: \( F_f = \mu_k F_N \), where \( F_N = mg\cos\theta \) (normal force).

Step 2: Net Force Formula

Net force \( F_{net} \) is the sum of forces along the incline:

$$ F_{net} = mg\sin\theta - \mu_k mg\cos\theta $$

Step 3: Substitute Values

Given:

• \( m = 1.00 \, \text{kg} \)
• \( g = 9.8 \, \text{m/s}^2 \)
• \( \theta = 30^\circ \)
• \( \mu_k = 0.40 \)

First, calculate \( \sin 30^\circ = 0.5 \) and \( \cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.866 \).

Substitute into the formula:

$$ F_{net} = (1.00)(9.8)(0.5) - (0.40)(1.00)(9.8)(0.866) $$

Step 4: Compute Each Term

• Gravitational component: \( (1.00)(9.8)(0.5) = 4.9 \, \text{N} \)
• Frictional force: \( (0.40)(1.00)(9.8)(0.866) \approx 3.41 \, \text{N} \)

Step 5: Calculate Net Force

$$ F_{net} = 4.9 - 3.41 = 1.49 \, \text{N} \approx 1.5 \, \text{N} $$

Answer:

The net force is approximately \( \boldsymbol{1.5 \, \text{N}} \) (matches the given value in the simulation).