Magnitude in force of two blocks connected by rope.

[asdehaan]

Homework Statement

Two blocks are connected by a rope F (Vertical in the drawing, with force F pulling on a rope, block 1 attached to that rope, then there ia another rope off the bottom of block 1 that attaches to block 2). The block starts from a rest and moves upward a distance of 3.00m in 5.00s while this force is being applied. The upper block (1) has a mass of 4.50kg and the lower block (2) has a mass of 9.00kg.

What is the magnitude of force F?
What is the tension in the rope connecting the blocks?

Homework Equations

F=ma

The Attempt at a Solution

Acceleration = dv/dt. Since V is increasing at a constant rate, Vav is the same as V. a is the slope of the line of V(t). So:

Vav(3.00m-0m)/(5.00s-0s)=(3/5)(m/s)=.600(m/s)

a=(.600m/s)/(5.00m/s)=.120m/s^2

The forces in block 2 -> T(rope)=m2a
=9.00kg*.120m/s^2
=1.08N
The forces in block 1 -> F=T(rope)+m1a
=1.08N+4.50kg*.120m/s^2
=1.62N

And the force in the rope should be 1.08N.

If this is wrong, please point me in the right direction. It seems like the force of gravity should have some kind of an effect here, but I haven't used it.

Do I need to add the acceleration of gravity to my acceleration above to get the force of gravity to cancel out? So 9.80m/s^2+.120m/s^2=9.92m/s^2?

Thanks

 

[tiny-tim]

Welcome to PF!

Hi asdehaan! Welcome to PF!

Acceleration = dv/dt. Since V is increasing at a constant rate, Vav is the same as V. a is the slope of the line of V(t). So:

Vav(3.00m-0m)/(5.00s-0s)=(3/5)(m/s)=.600(m/s)

a=(.600m/s)/(5.00m/s)=.120m/s^2

Sorry, I don't understand what you're doing here …

you are given t and s and u (or v_i), and you want to find a …

so just use one of the standard constant acceleration equations.

 

[asdehaan]

Yeah it was late last night. I'm not sure what I was doing. ;) I got it all squared away with my teacher today. Thanks.
Btw - I was trying to calculate the acceleration from the average velocity.