Velocity and Forces: Find Time for 15 m/s Speed

[MechaMZ]

Homework Statement

time-dependent force, $$\vec{}F$$= (8.60i - 4.20tj) N (where t is in seconds) is applied to a 2.00 kg object initially at rest.

(a) At what time will the object be moving with a speed of 15.0 m/s?

The Attempt at a Solution

$$\vec{F}$$ = ma
$$\sqrt{8.6^2 + (4.20t)^2}$$ = 2a
a = $$\sqrt{8.6^2 + (4.20t)^2}$$ / 2

v_f = v_i +at
15 = 0 + [$$\sqrt{8.6^2 + (4.20t)^2}$$ / 2]t

this should be an easy question but I can't get the correct answer as 2.86s from the equations above, I can't figure out the reason. hope someone could enlighten me

 

[rl.bhat]

Check the last calculation.
Square both side.
225*4 = [8.6^2 + 4.2^2*t^2]*t^2
Solve for t.

 

[MechaMZ]

Check the last calculation.
Square both side.
225*4 = [8.6^2 + 4.2^2*t^2]*t^2
Solve for t.

Hi,

I think even we square both side, the answer is still incorrect. you will get 2.36s but the answer is 2.86s.

I think the problem is because the vf = vi +at, gives you the instantaneous velocity yet the question requires you to find speed.

does it make sense? please advise, thank you.

 

[kuruman]

Your method works only if the velocity is in the same direction as the acceleration at all times. This is not the case here because the x-component of the acceleration is constant while the y-component changes with time. You need to calculate v_x(t) and v_y(t) separately, then apply the Pythagorean theorem to get the speed as a function of time.

 

[Doc Al]

The problem is that vf = vi + at applies to constant acceleration, which this is not. (Integrate each component separately.)

 

[MechaMZ]

Your method works only if the velocity is in the same direction as the acceleration at all times. This is not the case here because the x-component of the acceleration is constant while the y-component changes with time. You need to calculate v_x(t) and v_y(t) separately, then apply the Pythagorean theorem to get the speed as a function of time.

Hi,

so what method should I use for the V_y?
however, I could still using v_f = v_i +at for V_x since it is a constant acceleration?
I was wondering the v in the equation above is for instantaneous velocity yet the question is asking about speed.

 

[Doc Al]

I was wondering the v in the equation above is for instantaneous velocity yet the question is asking about speed.

That's not an issue. The instantaneous speed is just the magnitude of the instantaneous velocity.

 

[MechaMZ]

Hi Doc,

How about the question below, was the elevator accelerating with a constant value or not? may i know how did you know the a is constant or not?

A 79.0 kg man stands on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.82 m/s in 0.500 s. It travels with this constant speed for the next 5.00 s. The elevator then undergoes a uniform acceleration in the negative y direction for 1.60 s and comes to rest.

any clue so that i could know the elevator is accelerating with a constant or not a constant value during the first 0.5s?

thank you

 

[Doc Al]

any clue so that i could know the elevator is accelerating with a constant or not a constant value during the first 0.5s?

Unless you are given information to the contrary, I would just assume that all accelerations are constant.

 

[MechaMZ]

but if i take 1.82/0.5 is the instantaneous acceleration or average acceleration or constant acceleration for each second from 0m/s to 1.82m/s?

 

[Doc Al]

but if i take 1.82/0.5 is the instantaneous acceleration or average acceleration or constant acceleration for each second from 0m/s to 1.82m/s?

If you assume that the acceleration is constant, then all three of those are the same! (Since Vf = Vi + at.)

 

[MechaMZ]

argh, how about this question, why all my assumptions are wrong regarding the acceleration, force and velocity.

The figure below shows the speed of a person's body as he does a chin-up. Assume the motion is vertical and the mass of the person's body is 68.0 kg.

http://img147.imageshack.us/img147/5012/p527.gif

(a) time zero
correct answer is 687 N, but i thought it should be 68x9.81=667.08

why?

 

[kuruman]

You don't say what the question is, but 68*9.81 N is the answer to "What is the person's weight?".

 

[Doc Al]

argh, how about this question, why all my assumptions are wrong regarding the acceleration, force and velocity.

You didn't tell us your assumptions, so we can't say what's wrong.

(a) time zero
correct answer is 687 N, but i thought it should be 68x9.81=667.08

Please post the exact question and your reasoning. I suspect the question asks for the force he exerts on the bar during the initial portion of the chin up. Since he's accelerating, the force must be greater than just his weight.

 

[MechaMZ]

Hi Doc,

the question is "Determine the force exerted by the chin-up bar on his body at the following times."

(a) time zero
correct answer is 687 N, but i thought it should be 68x9.81=667.08

I was thinking when the time is zero, the velocity(movement) is zero. so there is no any upwards resultant force. I assume the T-mg = 0, so T = mg - 68* 9.81.

please give me some clue on this, thank you =)

 

[kuruman]

The velocity may be zero, but his acceleration is not. Can you read the acceleration from the graph?

 

[Doc Al]

Hi Doc,

the question is "Determine the force exerted by the chin-up bar on his body at the following times."

(a) time zero
correct answer is 687 N, but i thought it should be 68x9.81=667.08

I was thinking when the time is zero, the velocity(movement) is zero. so there is no any upwards resultant force. I assume the T-mg = 0, so T = mg - 68* 9.81.

The question is somewhat poorly worded. When he's just hanging, before he starts his chinup, the force exerted by the chinup bar equals his weight. But you need to base your answer on the given diagram. (Note that they don't show the time before he starts pulling.) They are treating t = 0 as when he's already started pulling. A better way to phrase the question would have been: Determine the force exerted by the bar immediately after he starts the chinup.

 

[MechaMZ]

i see, so there is an acceleration.