Motion Near Earth's Surface - grade 11 physics (1 Viewer)

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1. The problem statement, all variables and given/known data
a) Cathy takes the bus home from work. In her hand she holds a 2.0kg cake box, tied together with a string. As she ascends the steps into the bus, the box accelerates upward at a rate of 2.5 m/s^2. What is the force exerted on the string?

2. Relevant equations


3. The attempt at a solution
= (2.0)(2.5)-(2.0)(9.8)
= 5-19.6
Therefore the force exerted on the string is 1.5x10N[upward]

-I am taking an independent physics course. I always do the question as best I can first, then I check this forum for verification, it helps so much. But for this question, it says that 24.6N is correct.
I followed the examples in my course book exactly so I don't know why it would be 24.6N.
I mean I understand how to get 24.6 but why would (2)(9.8)=19.6 turn negative if the box accelerates upward? Wouldn't it remain positive?
Or is 24.6 wrong? Confused.
Thanks in advance.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
your not taking into account that the box and the string act in equal and opposite directions. When you go up on an elevator, doesn't it feel as if you are of a greater mass? This concept is the exact same as the box and string.

you can see that 19.6+5 = 24.6N
Thank you :) that does make sense.
It's just that my course book shows:


and so that's what I've been following.
Could you possibly explain it expressed in this way?
In the original, you just did the math incorrectly: It's the F(up) - mg = ma. So F = ma + mg. In your coursebook, could you state the problem that it's listed for?
There is no question or example in my book that shows
F(up) - mg = ma
F = ma + mg

Only examples using
are used.

For example:
Carla carries a spring balance with her on a fishing trip. She catches a fish, weighs it on the scale and gets a reading of 5.0N
In the elevator Carla notices the scale no longer reads 55.0N. The elevator is accelerating upward at a rate of2.0m/s^2, what reading does she see?

=(5.61kg)(2.0m/s) - (-55.0N)
This following example is why I originally thought the answer -14.6N, was 14.6N[upward]:

A 50.0kg girl falls 3.0m from a loft into a pile of hay. She is brought to rest by the hay in a distance of 1.0m. What is the average force exerted by the hay?

Is broken up into first part of the fall and second part:
at the very end of the solution for this example it shows:
=(50)(-29.4m/s) - (50)(9.8m/s)
Therefore the hay exerts a force of 2.0x10^3[upward] on the girl.

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