Math Problem Statement

A boy and his sled have a combined mass of 65 kg. What is their acceleration as they start down an icy 22.6 incline with a coefficient of friction equal to 0.10?

Solution

To solve for the acceleration of the boy and his sled as they start down the incline, we will analyze the forces acting on the system and apply Newton's second law.

Given:

  • Mass (mm) = 65kg65 \, \text{kg}
  • Angle of incline (θ\theta) = 22.622.6^\circ
  • Coefficient of friction (μk\mu_k) = 0.100.10
  • Gravitational acceleration (gg) = 9.8m/s29.8 \, \text{m/s}^2

Step 1: Identify forces acting along and perpendicular to the incline.

  1. Gravitational force (FgF_g): Fg=mg=659.8=637NF_g = m \cdot g = 65 \cdot 9.8 = 637 \, \text{N}

  2. Component of gravitational force parallel to the incline (FparallelF_{\text{parallel}}): Fparallel=Fgsin(θ)=637sin(22.6)F_{\text{parallel}} = F_g \cdot \sin(\theta) = 637 \cdot \sin(22.6^\circ)

  3. Component of gravitational force perpendicular to the incline (FperpendicularF_{\text{perpendicular}}): Fperpendicular=Fgcos(θ)=637cos(22.6)F_{\text{perpendicular}} = F_g \cdot \cos(\theta) = 637 \cdot \cos(22.6^\circ)

  4. Frictional force (FfrictionF_{\text{friction}}): Ffriction=μkFperpendicular=0.10(637cos(22.6))F_{\text{friction}} = \mu_k \cdot F_{\text{perpendicular}} = 0.10 \cdot (637 \cdot \cos(22.6^\circ))


Step 2: Net force along the incline.

The net force (FnetF_{\text{net}}) is the difference between the downhill force (FparallelF_{\text{parallel}}) and the frictional force (FfrictionF_{\text{friction}}): Fnet=FparallelFfrictionF_{\text{net}} = F_{\text{parallel}} - F_{\text{friction}}


Step 3: Acceleration (aa) using Newton's second law.

From Newton's second law (Fnet=maF_{\text{net}} = m \cdot a): a=Fnetma = \frac{F_{\text{net}}}{m}


Calculations:

  1. Compute FparallelF_{\text{parallel}}: Fparallel=637sin(22.6)6370.384244.7NF_{\text{parallel}} = 637 \cdot \sin(22.6^\circ) \approx 637 \cdot 0.384 \approx 244.7 \, \text{N}

  2. Compute FperpendicularF_{\text{perpendicular}}: Fperpendicular=637cos(22.6)6370.924588.4NF_{\text{perpendicular}} = 637 \cdot \cos(22.6^\circ) \approx 637 \cdot 0.924 \approx 588.4 \, \text{N}

  3. Compute FfrictionF_{\text{friction}}: Ffriction=0.10588.458.8NF_{\text{friction}} = 0.10 \cdot 588.4 \approx 58.8 \, \text{N}

  4. Compute FnetF_{\text{net}}: Fnet=244.758.8185.9NF_{\text{net}} = 244.7 - 58.8 \approx 185.9 \, \text{N}

  5. Compute acceleration aa: a=Fnetm=185.9652.86m/s2a = \frac{F_{\text{net}}}{m} = \frac{185.9}{65} \approx 2.86 \, \text{m/s}^2


Final Answer:

The acceleration of the boy and his sled is approximately 2.86m/s22.86 \, \text{m/s}^2.


Would you like further details or clarification on any step?

Expanding Questions:

  1. How does changing the angle of the incline affect the acceleration?
  2. What would happen if the surface were frictionless?
  3. How is acceleration affected by the mass of the boy and sled combined?
  4. Can we calculate the time taken to cover a certain distance with this acceleration?
  5. What is the role of the coefficient of friction in determining the net force?

Tip:

Always resolve forces into components parallel and perpendicular to the surface when analyzing motion on an incline.

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Math Problem Analysis

Mathematical Concepts

Physics
Forces on an Incline
Newton's Second Law
Trigonometry
Kinetics

Formulas

Gravitational Force (Fg = m * g)
Parallel Component of Gravitational Force (F_parallel = Fg * sin(theta))
Perpendicular Component of Gravitational Force (F_perpendicular = Fg * cos(theta))
Frictional Force (F_friction = μ * F_perpendicular)
Net Force (F_net = F_parallel - F_friction)
Acceleration (a = F_net / m)

Theorems

Newton's Second Law of Motion

Suitable Grade Level

Grades 10-12