# Two blocks of power (2)

1. Oct 22, 2008

### soupastupid

1. The problem statement, all variables and given/known data

the force is given as F
the distance pushed is given as D

2. Relevant equations

W= F*d
F=ma
KE = 1/2 mv^2

3. The attempt at a solution

the blocks have the same kinetic energy

#### Attached Files:

• ###### power.bmp
File size:
276 KB
Views:
60
2. Oct 23, 2008

### Staff: Mentor

Correct. Since the same work (F*d) is performed, the resulting KE is the same.

3. Oct 23, 2008

### soupastupid

ok for part b it is asking how fast the lighter block is moving when compared to the big block.

im thinking

the F*D is the same

and KE is .5mv^2

and so

.5(4m)(v_L)^2 = .5m(v_S)^2

the m and .5 cancel out

so i have

4(V_L)^2=(V_S)^2

SQRT BOTH SIDE

and then

V_S = 2V_L

so therefore

the smaller box is moving twice as fast as the big box?

right?

4. Oct 23, 2008

### Staff: Mentor

Perfect!

5. Oct 23, 2008

### soupastupid

ok

now both the blocks have the same speed
and now have the same V

ok...

so i have to find out how far each box is pushed

and
work of larger box =

W_L= F*D_L
W_S=F*D_S

KE_L = .5(4M)(v)^2
KE_S = .5(M)(v)^2

and since W=chainge in KE (KE_initial is 0)

F*D_L = .5(4M)(v)^2
(v)^2 = (F*D_L) / .5(4M)

F*D_S = .5(M)(v)^2
(v)^2 = (F*D_S)/.5(M)

(F*D_S)/.5(M) = (F*D_L) / .5(4M)

.5 M and F cancel out

D_S = (D_L) / 4
D_L = 4D_S

the distance the larger box has to be pushed is 4 times larger than that of the smaller box?

#### Attached Files:

• ###### 2blockspartC.JPG
File size:
52.5 KB
Views:
52
6. Oct 23, 2008

Exactly!