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Get Full Access to University Physics - 13 Edition - Chapter 5 - Problem 24e
Get Full Access to University Physics - 13 Edition - Chapter 5 - Problem 24e

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# CP CALC A 5.00-kg crate is suspended from the end of a

ISBN: 9780321675460 31

## Solution for problem 24E Chapter 5

University Physics | 13th Edition

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University Physics | 13th Edition

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Problem 24E

CP CALC? A 5.00-kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t) is applied to the end of the rope, and the height of the crate above its initial position is given by y(t) = (2.80 m/s)t + (0.610 m/s3)t3. What is the magnitude of F when t = 4.00 s?

Step-by-Step Solution:
Step 1 of 3

Solution 24E dy =v (y)=2.8 m/s+(1.83 m/s )t 3 2 dt dvy 3 3 dt =a (y)=3.66 m/s +(1.83 m/s )t , At t=4 s, dvy 2 dt =a (y)=14.64 m/s Newton's law in y direction F-mg=ma 2 F=49 N+(5 kg)(14.64 m/s ) =122 N The force is greater than the weight hence the crate accelerates upwards.

Step 2 of 3

Step 3 of 3

##### ISBN: 9780321675460

The full step-by-step solution to problem: 24E from chapter: 5 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. Since the solution to 24E from 5 chapter was answered, more than 473 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: University Physics, edition: 13. University Physics was written by and is associated to the ISBN: 9780321675460. The answer to “CP CALC? A 5.00-kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t) is applied to the end of the rope, and the height of the crate above its initial position is given by y(t) = (2.80 m/s)t + (0.610 m/s3)t3. What is the magnitude of F when t = 4.00 s?” is broken down into a number of easy to follow steps, and 61 words. This full solution covers the following key subjects: crate, rope, end, magnitude, Force. This expansive textbook survival guide covers 26 chapters, and 2929 solutions.

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