Math Problem Statement

Two forces, F1 and F2, act on the 7.00-kg block shown in the drawing. The magnitudes of the forces are F1 = 59.2 N and F2 = 20.5 N. Take the positive direction to be to the right. Find the horizontal acceleration of the block, including sign.

Solution

To find the horizontal acceleration of the block, we can apply Newton's second law F=maF = ma. We'll need to:

  1. Break down the forces into their horizontal components.
  2. Sum up all horizontal forces.
  3. Use the total horizontal force to find the acceleration.

Given:

  • F1=59.2NF_1 = 59.2 \, \text{N}
  • F2=20.5NF_2 = 20.5 \, \text{N}
  • Mass of the block m=7.00kgm = 7.00 \, \text{kg}
  • Angle θ=70\theta = 70^\circ for F1F_1
  • Positive direction is to the right.

Step 1: Resolve F1F_1 into components

Since F1F_1 is applied at an angle of 70° from the horizontal, we can resolve it into horizontal and vertical components:

  • Horizontal component of F1F_1: F1x=F1cos(70)F_{1x} = F_1 \cos(70^\circ)
  • Vertical component of F1F_1 (not relevant for horizontal acceleration): F1y=F1sin(70)F_{1y} = F_1 \sin(70^\circ)

Step 2: Horizontal forces

  • F1F_1 has a horizontal component F1xF_{1x} to the right.
  • F2F_2 is horizontal and points to the left (so it's negative when summing forces).

Thus, the total horizontal force is: Ftotal=F1xF2F_{\text{total}} = F_{1x} - F_2 Substituting the expressions: Ftotal=F1cos(70)F2F_{\text{total}} = F_1 \cos(70^\circ) - F_2

Step 3: Apply Newton's second law to find acceleration

Ftotal=maF_{\text{total}} = ma a=Ftotalma = \frac{F_{\text{total}}}{m}

Now, let's calculate the values.The horizontal acceleration of the block is approximately 0.036m/s2-0.036 \, \text{m/s}^2. The negative sign indicates that the acceleration is to the left, opposite to the positive direction defined as to the right.

Would you like more details or have any other questions?

Here are some related questions:

  1. How does the vertical component of F1F_1 affect the system?
  2. What would happen if the angle of F1F_1 were smaller?
  3. How would the acceleration change if the block had more mass?
  4. What are the forces acting on the block in the vertical direction?
  5. How would the problem change if there were friction?

Tip: In problems with forces acting at angles, always resolve the forces into components before applying Newton's second law.

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Math Problem Analysis

Mathematical Concepts

Newton's Second Law
Vector Resolution
Trigonometry

Formulas

F = ma
Horizontal component: F1x = F1 * cos(θ)
Total force: Ftotal = F1x - F2
Acceleration: a = Ftotal / m

Theorems

Newton's Second Law of Motion

Suitable Grade Level

Grades 10-12