A 5.25 kg crate is suspended from the end of a short vertical rope of negligible

Question

A 5.25 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)=(2.80m/s)t+(0.61 m/s3 )t3

Part A

What is the magnitude of the force F when 4.50 s ?

Express your answer with the appropriate units.

SubmitMy AnswersGive Up

A 5.25 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)=(2.80m/s)t+(0.61 m/s3 )t3

Part A

What is the magnitude of the force F when 4.50 s ?

Express your answer with the appropriate units.

SubmitMy AnswersGive Up

Explanation / Answer

2 time differentiate gives acceleration

y(t)=(2.80m/s)t+(0.61 m/s3 )t3

so

a=6*0.61*t=3.66*t

Newton's 2nd Law

F=Ma=5.25*3.66*t

at t=4.5

F=Ma=5.25*3.66*4.5=86.4675 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.