macmade
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This is my own question, so sorry if the question might not be as clear.
During the Bench Press exercise, a person unracks a 200kg barbell and places it 46.5 cm above his/her chest. It took the person 1.06 seconds to lower the barbell to his/her chest at a constant speed. Then the person takes 0.92 seconds to get the bar 17 cm off his/her chest and 0.8 seconds to lockout the barbell at 46.5 cm. Calculate the amount of force it took for the entire rep.
Fnormal = Mass x (Acceleration + Gravity)
X(final) = X(start) + V(start)t + 1/2 at^2
a=accleration t=time
I think I solved the ascending part of the bench press but I don't know where to begin with the descending part.
The ascending phase is basically:
X(final) = X(start) + V(start)t + 1/2at^2
.17m = 0 + 0(0.92) + 1/2a(0.92)^2
a = 0.4017m/s^2
Fnormal = m(a+g)
Fnormal = 200(0.4017+9.81)
Fnormal = 2054.34 N
X(final) = X(start) + V(start)t + 1/2at^2
.295 = 0 + 0(0.8) + 1/2a(0.8)^2
a = 0.922m/s^2
Fnormal = m(a+g)
Fnormal = 2146.4 N
2146.4 + 2054.34 = 4200.74 N
I think that is the amount of force exerted during the ascending part of the bench press
Now I have no idea how to do the descending part but this is what I think. I am pretty sure the equations used for the descending phase are the same, since force and gravity will always be 200 kg and 9.81, the difficult part would be calculating the accleration. Since in descending, the bar is being "pulled" down to your chest, so that means gravity is working for the lifter. This is all I could get. Any help will be appreciated!
Homework Statement
During the Bench Press exercise, a person unracks a 200kg barbell and places it 46.5 cm above his/her chest. It took the person 1.06 seconds to lower the barbell to his/her chest at a constant speed. Then the person takes 0.92 seconds to get the bar 17 cm off his/her chest and 0.8 seconds to lockout the barbell at 46.5 cm. Calculate the amount of force it took for the entire rep.
Homework Equations
Fnormal = Mass x (Acceleration + Gravity)
X(final) = X(start) + V(start)t + 1/2 at^2
a=accleration t=time
The Attempt at a Solution
I think I solved the ascending part of the bench press but I don't know where to begin with the descending part.
The ascending phase is basically:
X(final) = X(start) + V(start)t + 1/2at^2
.17m = 0 + 0(0.92) + 1/2a(0.92)^2
a = 0.4017m/s^2
Fnormal = m(a+g)
Fnormal = 200(0.4017+9.81)
Fnormal = 2054.34 N
X(final) = X(start) + V(start)t + 1/2at^2
.295 = 0 + 0(0.8) + 1/2a(0.8)^2
a = 0.922m/s^2
Fnormal = m(a+g)
Fnormal = 2146.4 N
2146.4 + 2054.34 = 4200.74 N
I think that is the amount of force exerted during the ascending part of the bench press
Now I have no idea how to do the descending part but this is what I think. I am pretty sure the equations used for the descending phase are the same, since force and gravity will always be 200 kg and 9.81, the difficult part would be calculating the accleration. Since in descending, the bar is being "pulled" down to your chest, so that means gravity is working for the lifter. This is all I could get. Any help will be appreciated!