Problem involving Newton's 2nd law of motion

In summary: This is because the velocity and time are linked; doubling the acceleration will also double the velocity. So, if we use the same units as before, the final equation becomes\begin{aligned*}v^2 &= v_0^2 + 2 at^2 \\2 at^2 &= 625\; \text{m}^2/\text{s}^2 \\x &= \frac{312.5\; \text{m}^2/\text{s}^2}{a} \\ & \\v &= v_0 + 2 at^2 \\at^2 &= 625\; \text{m}^2/\text
  • #1
vkrock
8
0

Homework Statement



An arrow, starting from rest, leaves the bow with a speed of 25.0m/s. If the average force exerted on the arrow by the bow were doubled, all else remaining the same, with what speed would the arrow leave the bow?



Homework Equations



F = ma


The Attempt at a Solution



For a force, F, the launch velocity of the arrow was 25.0m/s. Acceleration is F/m and multiplying F/m by a time t gives velocity at time t. So I re-wrote the first scenario as 25.0m/s = t/m * F. In the second scenario the force is doubled (2F), so I re-wrote the velocity as Xm/s = t/m * 2F. Since t and m remain the same for both scenarios, is the velocity of the 2nd scenario simply that of the 1st scenario's but doubled (50m/s)?
 
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  • #2
all else remaining the same
One has to ask, what does that mean? i.e. the time remains the same or the displacement of the draw.

Doubling the force on the same mass would double the acceleration, and for the same the displacement, that would reduce the time of the launch.
 
  • #3
more detail

By all else remaining constant, it means that the arrow's displacement (x) remains the same. That seemed to be the general class consensus. The acceleration doubles when the force is doubled, that makes sense, but the next step (actually finding the velocity of such an arrow) is giving me some trouble.

What I've done is try to solve some of the kinematic equations to get expressions for displacement and time from which I can calculate velocity; I am running into troubles, because for the first arrow (acceleration = a) I can't get the velocity to come out as 25m/s, i always get half that.

This is the approach I am taking:

First arrow: V0 = 0m/s, V = 25m/s, Acceleration = a, Displacement = x

Finding x in terms of a:

V^2 = V0^2 + 2ax
625 = 2ax
312.5/a = x

Finding t in terms of a:

V = V0 + at
25 = at
25/a = t

Finding 1st a using new expression for and t:

V = V0 + at
25 = 312.5/a * 25/a
25 = 7812.5/a^2
a^2 = 7812.5/25
a = sqrt(7812.5/25) = 17.68 m/s2

check: V = 17.68 * 25/17.68 = 25m/s

Using initial acceleration*2 to find velocity after a doubled force (acceleration)

V = V0 + at
V = (17.68*2) * 25/(17.68*2)
V = 35.36 * 25/35.36
V = 25

The same velocity? The expressions seem to work ok for the 1st arrow's conditions, 17.68m/s acceleration matches the 25m/s launch velocity given in the problem. Can anyone help me figure out what's going wrong?
 
  • #4
vkrock said:
Finding 1st a using new expression for and t:

V = V0 + at
25 = 312.5/a * 25/a
Why did you substitute the expression for x as the value for a? If you had kept the units you would have seen that what you did here is invalid. So what are the units?

vkrock said:
First arrow: V0 = 0m/s, V = 25m/s, Acceleration = a, Displacement = x

Good so far; you even have units on the initial velocity (0 m/s rather than 0).

Finding x in terms of a:

V^2 = V0^2 + 2ax
625 = 2ax
312.5/a = x

Finding t in terms of a:

V = V0 + at
25 = at
25/a = t

Now you are dropping units. This would be much better done as

[tex]\begin{aligned*}
v^2 &= v_0^2 + 2 a x \\
2 a x &= 625\; \text{m}^2/\text{s}^2 \\
x &= \frac{312.5\; \text{m}^2/\text{s}^2}{a} \\ & \\
v &= v_0 + at \\
at &= 25 \text{m/s} \\
t &= \frac{25\; \text{m/s}}{a}
\end{aligned*}[/tex]

There is no way to find a numerical solution for the acceleration a because you don't know the displacement x. You can, however, still determine what would happens when the acceleration doubles.
 
Last edited:

1. What is Newton's 2nd law of motion?

Newton's 2nd law of motion states that the acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to its mass. In simpler terms, the larger the force applied to an object, the greater its acceleration will be, and the greater the mass of the object, the smaller its acceleration will be.

2. What is the formula for Newton's 2nd law of motion?

The formula for Newton's 2nd law of motion is F = ma, where F represents the net force, m represents the mass of the object, and a represents the acceleration.

3. How is Newton's 2nd law of motion used in real life?

Newton's 2nd law of motion is used in a variety of real-life situations, such as calculating the force needed to move an object, understanding the motion of objects in space, and designing vehicles and machines.

4. What are some common examples of Newton's 2nd law of motion?

Some common examples of Newton's 2nd law of motion include a person pushing a shopping cart, a car accelerating on a highway, a rocket launching into space, and a person jumping off a diving board.

5. How does mass affect the acceleration of an object according to Newton's 2nd law of motion?

According to Newton's 2nd law of motion, the acceleration of an object is inversely proportional to its mass. This means that as the mass of an object increases, its acceleration decreases, and vice versa.

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