Answer:

Explanation:
The acceleration of the block can be found by the kinematics equations:

Since the plane is frictionless, the only force acting on the block along the motion of the block is its weight.

0.0605J is your answer. Use the formula KE=1/2mv^2
Complete Question
The complete question is shown on the first uploaded image
Answer:
The workdone is 
Explanation:
From the question we are told that
The initial Volume is 
The final volume is 
The external pressure is
Generally the change in volume is

Substituting values we have


Generally workdone is mathematically represented as

W is negative because the working is done on the environment by the system which is indicated by volume increase
Substituting values


Now 
Therefore 
