Question 1

6 identical books are lying on a desktop. a tidy student decides to stack the books one on top of the other. if students do 11 J of work width of the spine of each book is 2.5 what is the mass

Question 2

$\begin{gathered} W_T=11J \\ 6\text{ books} \\ m=\text{?} \\ y=2.5\text{ cm= 0.025m} \\ In\text{ this case the work is due to the }weight\text{ of each book. But one book} \\ \text{wont moved because it will be on the bottom } \\ W=\text{mgy} \\ g=9.81m/s^2 \\ \text{For the second book} \\ W=m(9.81m/s^2)(0.025m)=m(0.24525m^2/s^2\text{)} \\ \text{For the third book} \\ W=m(9.81m/s^2)(0.025m+0.025m)=m(0.4905m^2/s^2) \\ \text{For the fourth book} \\ W=m(9.81m/s^2)(0.025m+0.025m+0.025m)=m(0.73575m^2/s^2) \\ \text{For the fifth book} \\ W=m(9.81m/s^2)(0.025m+0.025m+0.025m+0.025m)=m(0.981m^2/s^2) \\ \text{For the sixth book} \\ W=m(9.81m/s^2)(0.025m+0.025m+0.025m+0.025m+0.025m)=m(1.22625m^2/s^2) \\ W_T=m(0.24525m^2/s^2\text{)+}m(0.4905m^2/s^2)+m(0.73575m^2/s^2)+m(0.981m^2/s^2)+m(1.22625m^2/s^2) \\ W_T=m(3.67874m^2/s^2) \\ 11J=m(3.67874m^2/s^2) \\ \text{Solving m} \\ m=(11J)/(3.67874m^2/s^2) \\ m=2.99\text{ kg}\approx3\operatorname{kg} \\ \text{The mass of each book is 3kg} \end{gathered}$

0 Comments

Please log in or register to add a comment.

Please log in or register to answer this question.

1 Answer

0 Comments

Please log in or register to add a comment.

Other Questions