1. Aug 29, 2013

### brenfox

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
t-w = -ma

3. 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?

2. Aug 29, 2013

### janhaa

The tension in the rope is:

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

or...?

3. Aug 29, 2013

### 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:

4. Aug 29, 2013

### 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.

5. Aug 29, 2013

### CAF123

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.