Answer:
A) T1 = 269.63 K 
T2 = 192.59 K
B) W = -320 KJ 
Explanation:
We are given;
Initial volume: V1 = 7 m³
Final Volume; V2 = 5 m³
Constant Pressure; P = 160 KPa
Mass; m = 2 kg
To find the initial and final temperatures, we will use the ideal gas formula;
T = PV/mR
Where R is gas constant of helium = R = 2.0769 kPa.m/kg
Thus;
Initial temperature; T1 = (160 × 7)/(2 × 2.0769) = 269.63 K
Final temperature; T2 = (160 × 5)/(2 × 2.0769) = 192.59 K
B) world one is given by the formula;
W = P(V2 - V1) 
W = 160(5 - 7)
W = -320 KJ
 
        
             
        
        
        
As we know that KE and PE is same at a given position
so we will have as a function of position given as

also the PE is given as function of position as

now it is given that
KE = PE
now we will have




so the position is 0.707 times of amplitude when KE and PE will be same
Part b)
KE of SHO at x = A/3
we can use the formula

now to find the fraction of kinetic energy



now since total energy is sum of KE and PE
so fraction of PE at the same position will be


 
        
             
        
        
        
I’m not sure sorry I really wish I could help
        
             
        
        
        
Potential energy at top:
PE = mgh
PE = 40 x 9.81 x 12
P.E = 4,708.8 J
Kinetic energy at bottom:
KE = 1/2 mv²
KE = 1/2 x 40 x 10²
K.E = 2,000 J
P.E = K.E + Frictional losses
Frictional losses = 4708 - 2000
Frictional losses = 2708 J
The answer is D.