Wednesday, October 21, 2020

407 A crate of mass 9.8 kg is pulled

A crate of mass 9.8 kg is pulled - Physics

ChemistryExplain daily providing Q&A content “#407 A crate of mass 9.8 kg is pulled" in Physics, Ap physics 1 practice test, Best colleges for physics, Best physics books, Best physics schools

ChemistryExplain “#407 A crate of mass 9.8 kg is pulled" in Physics, Ap physics 1 practice test, Best colleges for physics, Best physics books, Best physics schools
Get the Free Online Chemistry Q&A Questions And Answers with explain. To crack any examinations and Interview tests these Chemistry Questions And Answers are very useful. Here we have uploaded the Free Online Chemistry Questions. Here we are also given the all chemistry topic.

 ChemistryExplain team has covered all Topics related to inorganic, organic, physical chemistry, and others So, Prepare these Chemistry Questions and Answers with Explanation Pdf.

For More Chegg Questions

Free Chegg Question

A crate of mass 9.8 kg is pulled up a rough incline with an initial speed of 1.44 m/s. The pulling force is 90 N parallel to the incline, which makes an angle of 21.0° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 5.10 m.

(a) how much work is done by the gravitational force on the crate?

(b) Determine the increase in internal energy of the crate-incline system owing to friction.

(c) How much work is done by the 90-N force on the crate?

(d) What is the change in kinetic energy of the crate?

(e) What is the speed of the crate after being pulled 5.10 m?

Free Chegg AnswerFor More Chemistry Notes and Helpful Content Subscribe Our YouTube Chanel - Chemistry Explain  

Free Chegg Answer

Here ,

m = 9.8 Kg

u = 1.44 m/s

F = 90 N

theta = 21 degree

uk = 0.40

distance ,d = 5.1 m

a)

work done by gravitational force = - m * g * d * sin(theta)

work done by gravitational force = -9.8 * 9.8 * 5.1 * sin(21)

work done by gravitational force = -175.5 J

b)

increase in internal energy = uk * m * g * cos(theta) * d

increase in internal energy = 0.40 * 9.8 * 9.8 * cos(21) * 5.1

increase in internal energy = -182.91 J

c)

work done by the 90 N force = 90 * 5.1

work done by the 90 N force = 459 J

d)

change in kinetic energy = net work done

change in kinetic energy = -175.5 - 182.91 + 459

change in kinetic energy = 100.6 J

Labels: , ,

0 Comments:

Post a Comment

Subscribe to Post Comments [Atom]

<< Home