Math Problem Statement

Problem 5: (7% of Assignment Value) A horizontal force, F1 = 55 N, and a force, F2 = 10.3 N acting at an angle of θ to the horizontal, are applied to a block of mass m = 3.7 kg. The coefficient of kinetic friction between the block and the surface is μk = 0.2. The block is moving to the right.

Randomized Variables F1 = 55 N F2 = 10.3 N m = 3.7 kg

  • Part (a) ✔ Solve numerically for the magnitude of the normal force, FN in Newtons, that acts on the block if θ = 30°. FN = 41.45 ✔ Correct!

    Part (b) Solve numerically for the magnitude of acceleration of the block, a in m/s2, if θ = 30°. a =

Solution

To solve for the magnitude of acceleration aa of the block, we need to analyze the forces acting on it and apply Newton's Second Law.

Step 1: Analyze the Forces

  • Force F1F_1: This is a horizontal force of 55 N.

  • Force F2F_2: This force has a magnitude of 10.3 N and is applied at an angle θ=30\theta = 30^\circ to the horizontal. We can break this force into its horizontal and vertical components:

    • Horizontal component of F2F_2: F2x=F2cos(θ)F_{2x} = F_2 \cdot \cos(\theta)
    • Vertical component of F2F_2: F2y=F2sin(θ)F_{2y} = F_2 \cdot \sin(\theta)
  • Frictional Force: The frictional force FfrictionF_{\text{friction}} opposes the motion and can be calculated using the coefficient of kinetic friction μk=0.2\mu_k = 0.2 and the normal force FNF_N (which was previously calculated as 41.45 N): Ffriction=μkFNF_{\text{friction}} = \mu_k \cdot F_N

  • Normal Force FNF_N: The normal force was already determined as 41.45 N in part (a).

Step 2: Newton’s Second Law

To find the acceleration, we apply Newton’s Second Law in the horizontal direction. The forces in the horizontal direction are:

  • F1F_1
  • The horizontal component of F2F_2, F2xF_{2x}
  • The frictional force, FfrictionF_{\text{friction}}

The net force in the horizontal direction is: Fnet=F1+F2xFfrictionF_{\text{net}} = F_1 + F_{2x} - F_{\text{friction}}

Using Newton’s Second Law: Fnet=maF_{\text{net}} = m \cdot a Thus, the equation for acceleration is: a=Fnetma = \frac{F_{\text{net}}}{m}

Step 3: Calculation

Let’s calculate each of the components:

  • Horizontal component of F2F_2: F2x=10.3Ncos(30)=10.3N0.866=8.91NF_{2x} = 10.3 \, \text{N} \cdot \cos(30^\circ) = 10.3 \, \text{N} \cdot 0.866 = 8.91 \, \text{N}

  • Frictional force: Ffriction=0.241.45N=8.29NF_{\text{friction}} = 0.2 \cdot 41.45 \, \text{N} = 8.29 \, \text{N}

Now, we can find the net force: Fnet=55N+8.91N8.29N=55N+0.62N=55.62NF_{\text{net}} = 55 \, \text{N} + 8.91 \, \text{N} - 8.29 \, \text{N} = 55 \, \text{N} + 0.62 \, \text{N} = 55.62 \, \text{N}

Finally, using Newton’s Second Law to find the acceleration: a=55.62N3.7kg=15.03m/s2a = \frac{55.62 \, \text{N}}{3.7 \, \text{kg}} = 15.03 \, \text{m/s}^2

Final Answer:

The magnitude of the acceleration is approximately: a=15.03m/s2a = 15.03 \, \text{m/s}^2


Would you like further details or clarifications on this solution? Here are a few related questions you could explore:

  1. How would the acceleration change if the angle θ\theta were increased to 45°?
  2. What would happen to the normal force if the coefficient of friction were increased to 0.3?
  3. How can we calculate the work done by the frictional force on the block?
  4. What would the acceleration be if the block had a different mass, say 5 kg?
  5. How can the net force be affected if the direction of F2F_2 is reversed?

Tip: When calculating frictional forces, always ensure you're using the correct normal force, as it directly influences the frictional interaction between surfaces.

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Math Problem Analysis

Mathematical Concepts

Newton's Second Law
Friction
Vector Decomposition
Kinematics

Formulas

F_{2x} = F_2 * cos(θ)
F_{2y} = F_2 * sin(θ)
F_{friction} = μk * F_N
F_{net} = F_1 + F_{2x} - F_{friction}
a = F_{net} / m

Theorems

Newton's Second Law

Suitable Grade Level

Grades 11-12