- #1
Seiya
- 43
- 1
Two blocks with masses 4.0kg and 8.0 kg are connected by a string and slide down a 30 degree inclined plane. The coefficient of kinetic friction between the 4.0 kg block and the plane is 0.25; that between the 8.0 kg block and the plane is 0.35
a) calculate the acceleration of each block
This part is easy i got it right, don't even need anyone to check it... 4 kg = 2.775 , 8 kg =1.4750
used for both blocks : mgx-ff=ma
b) calculate the tension in the string
I said since the bottom block accelerates faster it will create a tension which means both blocks will be connected by a rigid string with a tension in it... so this means i could solve them as a system ... this is where i need someone to verify i did it right or if not correct me...
i said that
(mgx1+mgx2)-(ff1+ff2) = (m1+m2)a
They are connected so they will have a new a in common for both which comes out to 1.9 .
So now i can find the tension by using either box and the first equations trying to find their acceleration alone... instead now i have
(for the bottom box)
mgx-ff-T=m(1.9)
So i get T = 3.5N
if i use this t in a Newtons second law equation for the top one ill get 1.9 acceleration for it as well...
Can anyone let me know if i did any mistakes please? Thank you :) I really appreciate it
a) calculate the acceleration of each block
This part is easy i got it right, don't even need anyone to check it... 4 kg = 2.775 , 8 kg =1.4750
used for both blocks : mgx-ff=ma
b) calculate the tension in the string
I said since the bottom block accelerates faster it will create a tension which means both blocks will be connected by a rigid string with a tension in it... so this means i could solve them as a system ... this is where i need someone to verify i did it right or if not correct me...
i said that
(mgx1+mgx2)-(ff1+ff2) = (m1+m2)a
They are connected so they will have a new a in common for both which comes out to 1.9 .
So now i can find the tension by using either box and the first equations trying to find their acceleration alone... instead now i have
(for the bottom box)
mgx-ff-T=m(1.9)
So i get T = 3.5N
if i use this t in a Newtons second law equation for the top one ill get 1.9 acceleration for it as well...
Can anyone let me know if i did any mistakes please? Thank you :) I really appreciate it