Question

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

Answer 1

tension = 2.5 g = 2.5 * 9.81 Newtons

0.2 M g = 2.5 g
M = 2.5 / 0.2 = 12.5 kg

Answer 2

I need the answers and steps

Answer 3

M=2.5 /0.2=12.5 kg

Answer 4

No response

Answer 5

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

Answer 6

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