HCV-
newton's laws of motion chapter
exercise q no.16 and 23.
i m not able to draw diagrams which r necessary for these problems.
so plz manage from HCV
1123.
T - 0.3g = 0.3a
0.6g - T = 0.6a
0.6g -0.3g -0.3a = 0.6a
0.3g =0.9a
a = 10/3
Therefore velocity of m1 after two seconds will be
v = 10/3*2
v = 20/3
Since m2 is stopped acceleration on m1 will be -g
Time before it will become tight will be t
gt = 20/3
t = 2/3
Total time = 2/3
1116.
Consider from a inertial point of view
Forces on the elevator are Upward of the elevator and downward of the mass
T - 1.5g = 1.5a
3g - T = 3a
3g - 1.5g = 4.5a
a = g/3
T = g/2 - 1.5g
T = 2g Force = 4g
Mass = 4kg
Force of the elevator = 4*g/10
Total force = 4g + 4g/10
Mass = 4 + 4/10
Mass= 4+0.4
Mass measured = 4.4kg
11Pseudo Force will be downwards rite?
And wat is the value of pseudo force is g/10 rite?
T -1.5g - 1.5g/10 = 1.5a
3g + 3g/10 - T = 3a
3g +3g/10 -1.5g -1.5g/10 = 3a + 1.5a
4.5a = 1.5g +1.5g/10
a = g/3 + g/30
T - 1.5g - 1.5g/10 = g/2 + g /20
T = 1.5 g + 1.5g/10 +g/2 + g /20
T/g = 30 + 3 + 10 + 1/20
T /g= 44/20
2T/g = 44/10
Mass = 4.4 kg
Got it?
