Question

A 1100-kg car sits at rest at the bottom of a 15-degree hill. The driver initiates the accelerator, and the engine pushes it up a 15-degree hill with a net force of 4500 N parallel to the hill. There is an effective 0.1 coefficient of kinetic friction. Use energy concepts to calculate the speed of the car after it has moved 40 m along the hill's surface.

Name: a 1100 kg car sits at rest at the bottom of a 15 degree hill the driver initiates the accelerator and the engine pushes it up a 15 degree hill with a net force of 4500 n parallel to the hill 97611
Uploaded: 2022-07-13T05:12:50-08:00
Duration: 1 min 55 s
Channel: Ankur S
Description: a 1100 kg car sits at rest at the bottom of a 15 degree hill the driver initiates the accelerator and the engine pushes it up a 15 degree hill with a net force of 4500 n parallel to the hill 97611

          A 1100-kg car sits at rest at the bottom of a 15-degree hill. The driver initiates the accelerator, and the engine pushes it up a 15-degree hill with a net force of 4500 N parallel to the hill. There is an effective 0.1 coefficient of kinetic friction. Use energy concepts to calculate the speed of the car after it has moved 40 m along the hill's surface.

Added by Sandra M.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Ankur S

07/13/2022

Step 1

81 m/s^2, and theta = 15 degrees. F_parallel = 1100 kg * 9.81 m/s^2 * sin(15 degrees) = 1100 * 9.81 * 0.2588 = 2857.428 N Show more…

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A 1100-kg car sits at rest at the bottom of a 15-degree hill. The driver initiates the accelerator, and the engine pushes it up a 15-degree hill with a net force of 4500 N parallel to the hill. There is an effective 0.1 coefficient of kinetic friction. Use energy concepts to calculate the speed of the car after it has moved 40 m along the hill's surface.