tension = 2.5 g = 2.5 * 9.81 Newtons
0.2 M g = 2.5 g
M = 2.5 / 0.2 = 12.5 kg
2.1 Two blocks of mass M kg and 2.5kg respectively are connected by a light, inextensible string. The string runs over a light, frictionless pulley ,The blocks are stationary
2.1.1 calculate the tension in a string
The coefficient of static friction between the unknown mass M and the surface of the table is 0.2.
2.1.2 Calculate the minimum value of M that'll prevent the blocks from moving
6 answers
I need the answers and steps
M=2.5 /0.2=12.5 kg
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Fnet=ma
T-fs=ma
24.5-fs=M (0)
-fs=-24.5
:fs=24.5N
Fs=0.2×(M)(9.8)
24.5=1.96M
M=24.5/1.96
M=12.5kg
T-fs=ma
24.5-fs=M (0)
-fs=-24.5
:fs=24.5N
Fs=0.2×(M)(9.8)
24.5=1.96M
M=24.5/1.96
M=12.5kg
Fnet=0
T1-Fg=0
T1=mg
T1=2,5(9,8)
T1 =24,5
Fnet =0
T1+(-fk)=0
24,5-0,2N=0
N=122,5
Fn=m(9,8)
122,5÷9,8=9,8m÷9,8
M=12,5
T1-Fg=0
T1=mg
T1=2,5(9,8)
T1 =24,5
Fnet =0
T1+(-fk)=0
24,5-0,2N=0
N=122,5
Fn=m(9,8)
122,5÷9,8=9,8m÷9,8
M=12,5