Instrument attached to wire - Newton's laws

[cdlegendary]

Homework Statement

A 5.10 kg instrument is hanging by a vertical wire inside a spaceship that is blasting off at the surface of the earth. This ship starts from rest and reaches an altitude of 290 m in 10.0 s with constant acceleration.

a.) Draw a free-body diagram for the instrument during this time.(Assume that the spaceship is accelerating upward. )

b.) Find the force that the wire exerts on the instrument.

Homework Equations

f=ma
v=d/t
a=v/t

The Attempt at a Solution

I already drew the FBD, and it's just 2 vectors, the tension and weight, tension upward with more magnitude, and weight downward with less magnitude than that of tension.

The part I'm having trouble is with part 2, shouldn't the tension just be the weight of the instrument attached to the wire + the force of acceleration due to the ship?

So, T = mg + ma

T = (9.8m/s^2)(5.1kg) + (2.9m/s^2)(5.1kg)

This seems to be wrong though. (I got the 2.9m/s^2 through the v = d/t and a = v/t equations.)

 

[Spinnor]

Homework Statement

A 5.10 kg instrument is hanging by a vertical wire inside a spaceship that is blasting off at the surface of the earth. This ship starts from rest and reaches an altitude of 290 m in 10.0 s with constant acceleration.

a.) Draw a free-body diagram for the instrument during this time.(Assume that the spaceship is accelerating upward. )

b.) Find the force that the wire exerts on the instrument.

Homework Equations

f=ma
v=d/t
a=v/t

The Attempt at a Solution

I already drew the FBD, and it's just 2 vectors, the tension and weight, tension upward with more magnitude, and weight downward with less magnitude than that of tension.

The part I'm having trouble is with part 2, shouldn't the tension just be the weight of the instrument attached to the wire + the force of acceleration due to the ship?

So, T = mg + ma

T = (9.8m/s^2)(5.1kg) + (2.9m/s^2)(5.1kg)

This seems to be wrong though. (I got the 2.9m/s^2 through the v = d/t and a = v/t equations.)

v = d/t is average velocity so this is not a relevant equation.

But distance, d = a*t^2/2 for constant acceleration.

I got a = 2.4 m/s^2 not the 2.9 you got.

 

[cdlegendary]

v = d/t is average velocity so this is not a relevant equation.

But distance, d = a*t^2/2 for constant acceleration.

I got a = 2.4 m/s^2 not the 2.9 you got.

Ah I see. I thought I could use the average velocity to get the acceleration. I guess not. Well, when I use the d = (at^2)/2 equation, I get the acceleration to be 5.8m/s^2, not 2.4. What am I doing wrong?

a = 2d/t^2 = 2(290)/(10^2)

 

[Spinnor]

Ah I see. I thought I could use the average velocity to get the acceleration. I guess not. Well, when I use the d = (at^2)/2 equation, I get the acceleration to be 5.8m/s^2, not 2.4. What am I doing wrong?

a = 2d/t^2 = 2(290)/(10^2)

My mistake.