Finding force from position equation

[omega5]

Homework Statement

A 4.50kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t)is applied to the end of the rope, and the height of the crate above its initial position is given by $y(t) = (2.80m/s)t +(0.61m/s^3)t^3$ What is the force at 4 seconds?

Homework Equations

$$F = ma$$

The Attempt at a Solution

Taking the double derivative to get an acceleration equation:
$y'(t) = (3*0.61) t^2 + 2.80$
$y''(t) = a(t) = (0.61*3*2) t = 3.66t$

$F = 4.50 kg * 3.66 * 4.00s = 65.9 N$

Although MasteringPhysics tells me my answer's wrong, it doesn't give any explanation.

 

[rude man]

Homework Statement

A 4.50kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t)is applied to the end of the rope, and the height of the crate above its initial position is given by $y(t) = (2.80m/s)t +(0.61m/s^3)t^3$ What is the force at 4 seconds?

Homework Equations

$$F = ma$$

The Attempt at a Solution

Taking the double derivative to get an acceleration equation:
$y'(t) = (3*0.61) t^2 + 2.80$
$y''(t) = a(t) = (0.61*3*2) t = 3.66t$

$F = 4.50 kg * 3.66 * 4.00s = 65.9 N$

Although MasteringPhysics tells me my answer's wrong, it doesn't give any explanation.

Your last equation says force = mass x acceleration x time. Is that what Newton said?

 

[omega5]

I thought 3.66t would give the acceleration since the original equation gives the jerk.

 

[gneill]

One issue is that you didn't carry the units of the constants through your differentiations. Thus your constant in the last equation should be 3.66 m/s³.

A second issue is that I don't see where the force due to gravity on the crate is taken into account.

 

[omega5]

Ah! Thank you very much. That was the missing piece.