/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 81 A bicyclist travels in a circle ... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 81

A bicyclist travels in a circle of radius \(25.0 \mathrm{~m}\) at a constant speed of \(9.00 \mathrm{~m} / \mathrm{s}\). The bicycle-rider mass is \(85.0 \mathrm{~kg} .\) Calculate the magnitudes of (a) the force of friction on the bicycle from the road and (b) the net force on the bicycle from the road.

Short Answer

Expert verified
The force of friction and the net force on the bicycle are both 275.4 N.

Step by step solution

01

Calculate the Centripetal Force Required

To keep the bicyclist moving in a circle, a centripetal force is required. This force can be calculated using the formula: \[ F_c = \frac{mv^2}{r} \]where:- \( m \) is the mass of the bicycle-rider system, \( m = 85.0 \text{ kg} \)- \( v \) is the speed of the bicycle, \( v = 9.00 \text{ m/s} \)- \( r \) is the radius of the circle, \( r = 25.0 \text{ m} \)Substitute the known values into the formula:\[ F_c = \frac{85.0 \times (9.00)^2}{25.0} \]Calculate the numerical value to find the centripetal force.
02

Compute Centripetal Force

Continuing the calculation from Step 1:\[ F_c = \frac{85.0 \times 81}{25.0} = \frac{6885}{25.0} \]\[ F_c = 275.4 \text{ N} \]The centripetal force required is \( 275.4 \text{ N} \). This is the force of friction acting on the bicycle from the road.
03

Assess the Net Force

In the context of circular motion without vertical acceleration, the net force on the bicycle is equivalent to the centripetal force required to maintain the circular path. Hence, the net force is also \( 275.4 \text{ N} \).

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Key Concepts

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

Centripetal Force
When a bicyclist travels in a circular path, they rely on a force that acts towards the center of the circle to keep them on their curved path. This force is called the centripetal force. It doesn't push the object outwards, but pulls it inwards towards the center of the circle.
The centripetal force is crucial to maintain circular motion. Without it, an object would move off in a straight line according to Newton鈥檚 first law of motion.

The formula for centripetal force is given as:
  • \( F_c = \frac{mv^2}{r} \)
where:
  • \( m \) is mass,
  • \( v \) is velocity,
  • and \( r \) is the radius of the circle.

In our example, with a bicycle-rider mass of 85 kg, a speed of 9 m/s, and a circle radius of 25 m, plugging these into the formula gives a centripetal force of 275.4 N. This force is what causes the bicyclist to maintain their curved path instead of moving in a straight direction.
Frictional Force
In circular motion, frictional force plays a significant role, especially when it involves contact between surfaces like a bicycle tire on the road. Friction is the resisting force that occurs when two surfaces interact.

For the bicyclist moving in a circle, the frictional force from the road acts as the centripetal force. This means it is the friction that provides the necessary force for the cyclist to stay on their circular path.

It's essential to note that for circular motion, the direction of friction is towards the center of the circle, not opposite to the direction of motion. This is why, in our scenario, the frictional force equals the centripetal force required, which is 275.4 N.
  • Without sufficient friction, the bike would skid out of the circle.
  • On a slippery surface, this force might be less than the required centripetal force, leading to loss of control.
Net Force
In physics, net force is the total force acting on an object due to all its interactions. It's the sum of all forces that determine how the motion of an object will be affected.

For the bicyclist in circular motion, the net force is directed towards the center of the circle, aligning with our centripetal force and frictional force.

Since there's no vertical acceleration or other opposing forces impacting horizontally, the net force in this context is simply the centripetal force value, calculated at 275.4 N. This net force ensures the bicycle continues its path without deviating off course.
  • Net force determines acceleration, as noted in Newton's Second Law.
  • For uniform circular motion, net force is consistently the centripetal force.
Thus, the net force keeps the bicycle in smooth and continuous circular motion.

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Most popular questions from this chapter

A block slides with constant velocity down an inclined plane that has slope angle \(\theta\). The block is then projected up the same plane with an initial speed \(v_{0}\). (a) How far up the plane will it move before coming to rest? (b) After the block comes to rest, will it slide down the plane again? Give an argument to back your answer.

A house is built on the top of a hill with a nearby slope at angle \(\theta=45^{\circ}\) (Fig. 6-55). An engineering study indicates that the slope angle should be reduced because the top layers of soil along the slope might slip past the lower layers. If the coefficient of static friction between two such layers is \(0.5\), what is the least angle \(\phi\) through which the present slope should be reduced to prevent slippage?

\(68 \mathrm{~kg}\) crate is dragged across a floor by pulling on a rope attached to the crate and inclined \(15^{\circ}\) above the horizontal. (a) If the coefficient of static friction is \(0.50\), what minimum force magnitude is required from the rope to start the crate moving? (b) If \(\mu_{k}=0.35\), what is the magnitude of the initial acceleration of the crate?

A \(110 \mathrm{~g}\) hockey puck sent sliding over ice is stopped in \(15 \mathrm{~m}\) by the frictional force on it from the ice. (a) If its initial speed is \(6.0 \mathrm{~m} / \mathrm{s}\), what is the magnitude of the frictional force? (b) What is the coefficient of friction between the puck and the ice?

A box of canned goods slides down a ramp from street level into the basement of a grocery store with acceleration \(0.75 \mathrm{~m} / \mathrm{s}^{2}\) directed down the ramp. The ramp makes an angle of \(40^{\circ}\) with the horizontal. What is the coefficient of kinetic friction between the box and the ramp?
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