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Consider the acceleration of a car on dry pavement, if there is no slipping. The axle moves at speed v, and the outside of the tire moves at a speed relative to the axle. The instantaneous velocity of the bottom of the tire is zero. How much work is done by the force exerted on the tire by the road? What is the source of the energy that increases the car’s translational kinetic energy?

Short Answer

Expert verified

The work done by the force exerted on the tire by the road is0 Joule.Theinternal energy of the car causespeeding a car and this will result ingaining translational kinetic energy.

Step by step solution

01

Identification of given data

  • The speed of a caroutside of theisVout=v
  • The instantaneous velocity of the bottom of the tire is Vout=0
02

Concept of the work done by the force

The work done by the force is determined by the multiplication of force and displacement.

03

Determination of the work done by the force

The work done by the force can be calculated as,

The motion of tires of a car is pure rolling. Therefore, the work done the road friction on the tires will be 0.Hence, the work done by the force exerted on the tire by the road is 0 Joule.

The engine of a car gives power to the wheel. The increase in the car's kinetic energy comes from the internal energy of the car, this will result in gaining translational kinetic energy.

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Most popular questions from this chapter

Consider the acceleration of a car on dry pavement, if there is no slipping. The axle moves at speed v, and the outside of the tire moves at a speed relative to the axle. The instantaneous velocity of the bottom of the tire is zero. How much work is done by the force exerted on the tire by the road? What is the source of the energy that increases the car’s translational kinetic energy?

A string is wrapped around a uniform disk of mass M and radius R. Attached to the disk are four low-mass rods of radius b, each with a small mass m at the end (Figure 9.63).

The apparatus is initially at rest on a nearly frictionless surface. Then you pull the string with a constant force F. At the instant when the center of the disk has moved a distance d, an additional length w of string has unwound off the disk. (a) At this instant, what is the speed of the center of the apparatus? Explain your approach. (b) At this instant, what is the angular speed of the apparatus? Explain your approach.

Two identical 0.4 kgblock (labeled 1 and 2) are initially at rest on a nearly frictionless surface, connected by an unstretched spring, as shown in the upper portion of Figure 9.59.

Then a constant force of 100 N to the right is applied to block 2 and at a later time the blocks are in the new positions shown in the lower portion of Figure 9.59.9.59. At this final time, the system is moving to the right and also vibrating, and the spring is stretched. (a) The following questions apply to the system modeled as a point particle. (i) What is the initial location of the point particle? (ii) How far does the point particle move? (iii) How much work was done on the particle? (iv) What is the change in translational kinetic energy of this system? (b) The following questions apply to the system modeled as an extended object. (1) How much work is done on the right-hand block? (2) How much work is done on the left-hand block? (3) What is the change of the total energy of this system? (c) Combine the results of both models to answer the following questions. (1) Assuming that the object does not get hot, what is the final value of Kvib+Uspringfor the extended system? (2) If the spring stiffness is 50 N/m, what is the final value of the vibrational kinetic energy?

By calculating numerical quantities for a multiparticle system. One can get a concrete sense of the meaning of the relationships p→sys=Mtotv→CMand Ktot=Ktrans+Krel. Consider an object consisting of two balls connected by a spring, whose stiffness is 400 N/m. The object has been thrown through the air and is rotating and vibrating as it moves. At a particular instant, the spring is stretched 0.3m, and the two balls at the ends of the spring have the following masses and velocities: 1:5kg.(8,14,0)m/s2:3kg(-5,9,0)m/s

(a)For this system, calculate p→sys. (b) Calculate v→CM(c) Calculate Ktot3. (d) Calculate Ktrans. (e) Calculate Krel. (f) Here is a way to check your result for Krel. The velocity of a particle relative to the center of mass is calculated by subtracting v→CMfrom the particle’s velocity. To take a simple example, if you’re riding in a car that’s moving with v→CM,x=20m/sand you throw a ball with v→CM,x=35m/s, relative to the car, a bystander on the ground sees the ball moving with vx=55m/sSo v→=v→CM=v→reland therefore we have=v→relv→=v→CMfor each mass and calculate the correspondingKrel. Compare with the result you obtained in part (e).

Question: Under what conditions does the energy equation for the point particle system differ from the energy equation for the extended system? Give two examples of such a situation. Give one example of a situation where the two equations look exactly alike.

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