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Chapter 7: Problem 34

(I) In pedaling a bicycle uphill, a cyclist exerts a downward force of 450\(\mathrm { N }\) during each stroke. If the diameter of the circle traced by each pedal is \(36 \mathrm { cm } ,\) calculate how much work is done in each stroke.

Short Answer

Expert verified
The work done in each stroke is approximately 508.94 Joules.

Step by step solution

01

Understanding Work Done

Work done is calculated by the formula: Work = Force x Distance. The force is exerted in the direction of the motion, and in this exercise the force is given as 450 N.
02

Converting Diameter to Radius

Since the problem gives us the diameter of the circle traced by the pedal, we need the radius to calculate the circumference. Recall that the radius is half of the diameter. So, the radius, \( r \), is \( \frac{36 \text{ cm}}{2} = 18 \text{ cm} \).
03

Calculating the Circumference of the Circle

Knowing the radius, the circumference of the circle can be calculated using the formula: \[ C = 2\pi r \]Substitute \( r = 18 \text{ cm} \):\[ C = 2\pi \times 18 \text{ cm} = 36\pi \text{ cm} \].
04

Converting Circumference to Meters

Make sure all measurements are in the same unit. Convert \( 36\pi \text{ cm} \) to meters:\[ 36\pi \text{ cm} = 0.36\pi \text{ m} \].
05

Calculating Work Done Per Stroke

Use the formula for work: \[ \text{Work} = \text{Force} \times \text{Distance} \]Substitute the force (450 N) and the distance (\( 0.36\pi \text{ m} \)) into the equation:\[ \text{Work} = 450 \times 0.36\pi \]\[ \text{Work} = 162\pi \text{ Joules} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Force in Bicycle Mechanics
When riding a bicycle, the force you exert plays a crucial role in pushing you forward, especially uphill. In our scenario, this force is applied downward during each pedaling stroke.
Force is measured in newtons (N), and it's a concept that describes the push or pull on an object resulting from the object's interaction with another object. In bicycle mechanics, the force must align with the direction of motion to overcome resistance like gravity when going uphill.
The force applied by the cyclist in the exercise is 450 N each time the pedals are pushed down. This force acts in a linear path along the circular movement of the pedal. Understanding how this force leads to movement requires knowledge of the work-energy principle, which explains how the applied force translates into actual forward movement and work done.
Understanding Circumference
The term circumference refers to the distance around the edge of a circle. It's an essential concept when calculating the distance a bicycle pedal travels in one complete rotation.
To find the circumference of a circle, we use the formula:
  • \( C = 2\pi r \)
Where \( r \) is the radius of the circle. For the bicycle pedal, given in the exercise, we know the diameter is 36 cm. The radius \( r \) is then half of this, which is 18 cm. Substituting this into the circumference formula gives us \( C = 36\pi \) cm.
This gives the total distance any point on the pedal travels in a full circle, helping us understand how far the foot will move with each pedal stroke.
Bicycle Mechanics and Efficiency
Bicycle mechanics involves understanding how different parts like the frame, gears, and wheels work together. One critical aspect is how force applied on the pedals translates into the bicycle moving forward.
The efficiency of this system depends on how well the force is transformed into motion. It means reducing losses due to friction and ensuring maximum transfer of energy from the pedals to the motion of the wheels.
Essential elements:
  • Chains and gears help modify the force transferred to the wheels ensuring optimum power use.
  • The design of the wheel and tire reduces resistance and optimizes the transfer of force.
  • Rider position and technique also influence efficiency, allowing better application of force directly downward during pedaling.
In summary, understanding these aspects will help you appreciate the physics behind riding a bicycle and how mechanical designs maximize efficiency.

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Most popular questions from this chapter

A hammerhead with a mass of \(2.0 \mathrm{~kg}\) is allowed to fall onto a nail from a height of \(0.50 \mathrm{~m}\). What is the maximum amount of work it could do on the nail? Why do people not just "let it fall" but add their own force to the hammer as it falls?

An airplane pilot fell \(370 \mathrm{~m}\) after jumping from an aircraft without his parachute opening. He landed in a snowbank, creating a crater \(1.1 \mathrm{~m}\) deep, but survived with only minor injuries. Assuming the pilot's mass was \(88 \mathrm{~kg}\) and his terminal velocity was \(45 \mathrm{~m} / \mathrm{s}\), estimate: \((a)\) the work done by the snow in bringing him to rest; \((b)\) the average force exerted on him by the snow to stop him; and ( \(c\) ) the work done on him by air resistance as he fell. Model him as a particle.

Given vectors \(\mathbf{A}=-4.8 \mathbf{i}+6.8 \mathbf{j}\) and \(\mathbf{B}=9.6 \mathbf{i}+6.7 \mathbf{j}\), determine the vector \(\overrightarrow{\mathbf{C}}\) that lies in the \(x y\) plane perpendicular to \(\overrightarrow{\mathbf{B}}\) and whose dot product with \(\overrightarrow{\mathbf{A}}\) is \(20.0 .\)

We usually neglect the mass of a spring if it is small compared to the mass attached to it. But in some applications, the mass of the spring must be taken into account. Consider a spring of unstretched length \(\ell\) and mass \(M_{\mathrm{S}}\) uniformly distributed along the length of the spring. A mass \(m\) is attached to the end of the spring. One end of the spring is fixed and the mass \(m\) is allowed to vibrate horizontally without friction (Fig. \(7-30\) ). Each point on the spring moves with a velocity proportional to the distance from that point to the fixed end. For example, if the mass on the end moves with speed \(v_{0}\), the midpoint of the spring moves with speed \(v_{0} / 2 .\) Show that the kinetic energy of the mass plus spring when the mass is moving with velocity \(v\) is $$ K=\frac{1}{2} M v^{2} $$ where \(M=m+\frac{1}{3} M_{\mathrm{S}}\) is the "effective mass" of the system. [Hint: Let \(D\) be the total length of the stretched spring. Then the velocity of a mass \(d m\) of a spring of length \(d x\) located at \(x\) is \(v(x)=v_{0}(x / D) .\) Note also that $$ d m=d x\left(M_{\mathrm{S}} / D\right) $$.

A cyclist starts from rest and coasts down a 4.0^ \circ hill. The mass of the cyclist plus bicycle is 85\(\mathrm { kg }\) . After the cyclist has traveled \(250 \mathrm { m } ,\) (a) what was the net work done by gravity on the cyclist? (b) How fast is the cyclist going? Ignore air resistance.
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