Concept explainers
(a)
The constant friction force for the last
(a)
Answer to Problem 25AP
The constant friction force for the last
Explanation of Solution
Given info: The mass of the empty car is
The formula to calculate the energy by equation of motion is,
Here,
The formula of kinetic energy at start is,
Substitute
The formula of kinetic energy at final point is,
Substitute
The formula of the potential energy at start is,
Substitute
The formula of the potential energy at final point is,
Substitute
The formula of the energy wasted due to friction is,
Substitute
Substitute
Substitute
Conclusion:
Therefore, the constant friction force for the last
(b)
The highest speed reached by the car.
(b)
Answer to Problem 25AP
The highest speed reached by the car is
Explanation of Solution
Given info: The mass of the empty car is
The formula of kinetic energy at final point is,
The body is at the lowest point so the resistive force is taken upto that point only.
The formula of the energy wasted due to friction is,
Substitute
The formula to calculate the final velocity from equation (1) can be written as,
Substitute
Conclusion:
Therefore, the highest speed reached by the car is
(c)
The new values of friction force and highest speed when the weight is
(c)
Answer to Problem 25AP
The new value of friction force is
Explanation of Solution
Given info: The mass of the empty car is
The formula to calculate the
Substitute
To calculate the final velocity the equation (2) is,
Substitute
Conclusion:
Therefore, the new value of friction force is
(d)
The depth of the underground part of the ride.
(d)
Answer to Problem 25AP
The depth of the underground part of the ride is
Explanation of Solution
Consider
The formula of the potential energy at start is,
Substitute
The formula to calculate the depth is,
Substitute
Conclusion:
Therefore, the depth of the underground part of the ride is
(d)
The depth of the underground provided is feasible or not.
(d)
Answer to Problem 25AP
No, the depth of the underground provided is not feasible.
Explanation of Solution
No the depth of the underground tunnel provided is not feasible as the total length covered by the car is increased due to which the loss of energy due to friction is increased a lot as compared to the case when the underground path was not there. The non conventional forces on the car are increased due to the depth provided as the friction in the path of later
Conclusion:
Therefore, the depth of the underground provided is not feasible.
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Chapter 8 Solutions
Physics for Scientists and Engineers with Modern Physics
- In a truck-loading station at a post office, a small 0.200 kg package is released from rest at point A on a track that is one-quarter of a circle with radius 1.60 m. The size of the package is much less than 1.60 m, so the package can be treated as a particle. It slides down the track and reaches point B with a speed of 4.80 m/s. From point B, it slides on a level surface a distance of 3.00 m to point C, where it comes to rest. (a) What is the coefficient of kinetic friction on the horizontal surface? (b) How much work is done on the package by friction as it slides down the circular arc from A to B?arrow_forwardWhen a car is hit from behind, its passengers undergo sudden forward acceleration, which can cause a severe neck injury known as whiplash. During normal acceleration, the neck muscles play a large role in accelerating the head so that the bones are not injured. But during a very sudden acceleration, the muscles do not react immediately because they are flexible; most of the accelerating force is provided by the neck bones. Experiments have shown that these bones will fracture if they absorb more than 8.0 J of energy. (a) If a car waiting at a stoplight is rear-ended in a collision that lasts for 10.0 ms, what is the greatest speed this car and its driver can reach without breaking neck bones if the driver’s head has a mass of 5.0 kg (which is about right for a 70 kg person)? Express your answer in m/s and in mi/h. (b) What is the acceleration of the passengers during the collision in part (a), and how large a force is acting to accelerate their heads? Express the acceleration in m/s2…arrow_forwardYou push a .50kg block against a spring (k=3100 N/m),compressing it by .12m. The block is then released from rest and the spring pushes the block away. The spring and the block lose contact and the block collides with a second block of twice the mass. The two blocks slide together down a frictionless track consisting of a flat straightaway and a vertical, semi-circle of radius 40cm. What is the speed of the blocks when they have travelled halfway up the semicircle part of the track? What is the magnitude of the normal force on the two blocks at that same location?arrow_forward
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- Physics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage Learning