Question

Medals!!!


a 1.5kg book is sliding along a rough horizontal surface.At point a it is moving at 3.21m/s and at point b it slowed to 1.25m/s

a. how much work was done on the book between a and b
b. if -.750j of work is done on the book from b to c, how fast is it moving at point c?
c.how fast would it be moving at c if +.750j of work were done from b to c?


ill give medals!!!

Answer 1

Yes. Are you not given at least the coefficient of friction?

Answer 2

you assume its frictionless

Answer 3

Work done = total energy before - total energy after

Total energy = kinetic energy = 1/2 * mass * velocity^2

Between pt a and b, the velocity changes from 3.21 to 1.25. With the two eqns above, u can calculate the work done.

Answer 4

This is called work energy theorem
W=del k

Answer 5

can anyone explain further. Ive done a few problems similar but none like this.

Answer 6

Which part you have trouble following?

Answer 7

I dont understand what to plug in where. ive never deal with a change such as between a and b

Answer 8

What is the kinetic energy at point a?
What is the kinetic energy at point b?

Answer 9

a=2.4
b=.93 ?

Answer 10

kinetic energy = 1/2 * mass * velocity^2

show me how you get 2.4 plz?

Answer 11

.5*1.5*3.21 but I forgot to square 3.21

Answer 12

Yup. Try again plz?

Answer 13

7.72 and 1.17

Answer 14

So total energy before (a) - after (b) = work done...

Answer 15

so for the next two just use the same equation ?

Answer 16

Same idea but now the work done is given n u need to solve for velocity instead.

Answer 17

so -.750=.5*1.5v^2

Answer 18

-.750 = 1.17 - .5*1.5v^2

Answer 19

is that for b or c ?

Answer 20

Hahahaa good question: what do u think?

Answer 21

c

Answer 22

But the work done is -0.75 so the eqn is for b. of the question

Answer 23

thank you !

Answer 24

Welcome n thanx for the medal!