Question

Two pistons are in an enclosed volume with fluid in between them, such that
movement in one piston affects the location of the other piston, as shown in
the image below. If an explosion occurs in the piston on the left such that it
moves 1 m, what is the distance that the piston on the right will move
(assuming that friction can be neglected)? (Recall that work on an object is
equal to the force on the object times the distance it is moved, and that work
is conserved.)
Cross-sectional area - 25 mm
Cross-sectional area = 0,5 m
A. 2 m
B. 0.5 m
C. 1 m
D. 0.25 m

Answer 1

Answer:

The pressure in the fluid just below Piston 1 is

p_1=F_1/A_1=361\text{ Pa}.p

1

​

=F

1

​

/A

1

​

=361 Pa.

The pressure in the fluid just below Piston 2 is the same according to Pascal's law:

p_2=p_1=361\text{ Pa}.p

2

​

=p

1

​

=361 Pa.

The force being exerted on Piston 2 to keep the two pistons in equilibrium is

F_2=p_2A_2=330\text{ N}.F

2

​

=p

2

​

A

2

​

=330 N.

Explanation: