Calculate Tension of Rope for 0.5kg Mass Dropped 0.5m in 1.5s

[brenfox]

Homework Statement

a mass of 0.5kg is suspended from a flywheel. if the mass is released from rest and falls a distance of 0.5m in 1.5s, calculate : The tension in the rope. acceleration is 0.444ms-2.

Homework Equations

t-w = -ma

The Attempt at a Solution

t = m(ag)

t = 0.5(0.4444=9.81)

t = 5.127n

w = mg = 0.5x9.81 4.9n

so t-w 0.227n is rope tension.

Check by inserting numbers into equation of t-w = -ma

so: 5.127 - 4.905 = 0.5 x 0.222.

The last equation doesn`t marry up?

 

[janhaa]

The tension in the rope is:

T = m(g - a) = 0,5(9,81 - 0,44)

or...?

 

[abrewmaster]

If you had the mass just sitting there with no acceleration then it would be simply T=mg
Since it's accelerating you have to subtract the force acting on it from the acceleration that it is going at which is F=ma so therefore:
T=mg-ma=m(g-a)=0.5(9.81-0.444)=4.683N should be your answer?

 

[brenfox]

From the beginning

t-w = -ma

t=m(g-a) = 0.5(9.81-0.44)

= 4.683n so t = 4.683n

w=mg so w = 0.5x9.81 = 4.905n

S0 t-w = -ma

Checking... 4.683-4.905 = -0.5 x 0.444

Correct... i think.

 

[CAF123]

Homework Statement

a mass of 0.5kg is suspended from a flywheel. if the mass is released from rest and falls a distance of 0.5m in 1.5s, calculate : The tension in the rope. acceleration is 0.444ms-2.

Homework Equations

t-w = -ma

The Attempt at a Solution

t = m(ag)

t = 0.5(0.4444=9.81)

t = 5.127n

w = mg = 0.5x9.81 4.9n

so t-w 0.227n is rope tension.

Check by inserting numbers into equation of t-w = -ma

so: 5.127 - 4.905 = 0.5 x 0.222.

The last equation doesn`t marry up?

It is convenient to denote the positive direction in the direction of acceleration. In this case, the mass accelerates downwards so take down to be positive. This gives mg - T = ma. Solving gives T = m(g-a). It is difficult to see where you went wrong in your attempt.