Math Problem Statement

A child pulls a 15 kg sled containing a 5.0 kg dog along a straight path on a horizontal surface with a pulling force of 55 N at an angle of 20° above the horizontal. The coefficient of friction between the sled and the surface is 0.22. The tasks are: (a) Draw a free-body diagram for the system, (b) Calculate the normal force, (c) Calculate the acceleration of the system, (d) Calculate the work done by the pulling force as the sled moves 7.0 m, and (e) Sketch a speed vs. time graph when the dog rolls off.

Solution

Let's walk through the problem step by step.

Problem Overview:

A child pulls a sled containing a dog, and you are asked to:

  1. Draw a free-body diagram for the system.
  2. Calculate the normal force.
  3. Calculate the acceleration of the system.
  4. Calculate the work done by the child’s pulling force as the system moves.
  5. Analyze a case where the dog rolls off, and draw a graph of speed versus time for the sled.

Step-by-Step Breakdown:

(a) Free-Body Diagram:

The system consists of a sled and a dog being pulled. The forces acting on the system include:

  • Weight of the system (W): Downward due to gravity (sum of the sled and dog mass).
  • Normal force (N): Upward, perpendicular to the surface.
  • Friction force (f): Opposes the motion, proportional to the normal force.
  • Pulling force (F): At an angle of 20° above the horizontal.
  • Component of the pulling force: This has two parts:
    • Horizontal component: Fcos(20)F \cos(20^\circ)
    • Vertical component: Fsin(20)F \sin(20^\circ)

To draw the diagram:

  1. Represent the weight (W=(15+5)×9.8W = (15 + 5) \times 9.8 N).
  2. Show the normal force NN, upward.
  3. Show the friction force ff, opposing motion.
  4. Indicate the pulling force at a 20° angle, with its components labeled.

(b) Normal Force Calculation:

The normal force is affected by the pulling force's vertical component. The total vertical forces must balance because the sled is on a flat surface, so: N+Fsin(20)=(15+5)×9.8N + F \sin(20^\circ) = (15 + 5) \times 9.8 Solve for NN.

(c) Acceleration of the System:

Apply Newton’s Second Law in the horizontal direction. The net force is: Fnet=Fcos(20)fF_{\text{net}} = F \cos(20^\circ) - f Where f=μNf = \mu N is the friction force. Use Fnet=(mtotal)aF_{\text{net}} = (m_{\text{total}})a to solve for acceleration aa.

(d) Work Done:

The work done by the pulling force is given by: W=Fcos(20)×dW = F \cos(20^\circ) \times d where d=7.0d = 7.0 m is the distance.

(e) Speed-Time Graph:

After the dog rolls off, the system’s mass reduces, so the acceleration will change. The graph will show an increase in speed until the dog falls off (at time trt_r), followed by a steeper increase in speed (due to the smaller mass) as the sled continues to be pulled.

Would you like further details on any step or calculations?


Relative Questions:

  1. How do you calculate the horizontal and vertical components of the pulling force?
  2. Why does the normal force change when there is a vertical component of the pulling force?
  3. How does the frictional force depend on the normal force?
  4. What happens to the acceleration of the sled after the dog rolls off?
  5. How would the work done change if the coefficient of friction were different?

Tip: Always break down forces into their components when dealing with forces at an angle.

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Math Problem Analysis

Mathematical Concepts

Newton's laws of motion
Free-body diagrams
Kinematics
Work and energy
Forces and friction

Formulas

Normal force: N = (m_total * g) - F * sin(θ)
Net force: F_net = F * cos(θ) - f_friction
Acceleration: a = F_net / m_total
Work: W = F * cos(θ) * d

Theorems

Newton's second law (F = ma)
Work-energy theorem

Suitable Grade Level

Grades 9-12