Jane and John, with masses of 50 kg and 60 kg, respectively, stand on a frictionless surfact 10 m apart. John pulls on a rope that connects him to Jane, giving Jane an acceleration of 0.92 m/s^2 toward him. (a) What is John's acceleration? (b) If the pulling force is applied constantly, where will Jane and John meet?

Question

Jane and John, with masses of 50 kg and 60 kg, respectively, stand on a frictionless surfact 10 m apart. John pulls on a rope that connects him to Jane, giving Jane an acceleration of 0.92 m/s^2 toward him. (a) What is John's acceleration? (b) If the pulling force is applied constantly, where will Jane and John meet?

Answer 1

(a) What is John's acceleration?

Well, if Jane is accelerating towards John, then John must be accelerating towards Jane as well. You see, love can be a powerful force! So, let's call John's acceleration "x" (which I promise is not a variable for hugs and kisses). And since mass times acceleration equals force, we can use some math magic to find the solution.

Using Newton's second law, we can say that the force exerted on Jane is equal to mass times acceleration, or F = m*a. And since John is doing the pulling, the force applied to Jane is also the force applied by John. So, we have:

F = m*a
F = (60 kg)(x)

But we know that the force applied is also equal to the product of Jane's mass and her acceleration, F = (50 kg)(0.92 m/s^2).

Setting these two equal to each other, we get:

(60 kg)(x) = (50 kg)(0.92 m/s^2)

Now, we can solve for x:

x = (50 kg)(0.92 m/s^2) / 60 kg

x ≈ 0.767 m/s^2

So, John's acceleration is approximately 0.767 m/s^2. It seems like he's going to have to pick up the pace to catch up with Jane!

(b) If the pulling force is applied constantly, where will Jane and John meet?

Well, since they both have constant accelerations, we can use the kinematic equation to determine where they will meet. Remember, distance equals the initial velocity times time plus one-half the acceleration times time squared. In this case, John is the one pulling, so let's assume he's the one with the initial velocity of zero.

The distance between them is 10 m, and we need to find the time it takes for them to meet. Let's call that time "t".

Using the equation:

distance = initial velocity * time + (1/2) * acceleration * time^2

We have:

10 m = 0 * t + (1/2) * 0.92 m/s^2 * t^2

Simplifying, we get:

5t^2 = 10

Dividing both sides by 5, we have:

t^2 = 2

Taking the square root of both sides, we get:

t = √2

So, Jane and John will meet after approximately √2 seconds. That's the square root of 2, not the "scream like a maniac" square root of 2! Remember, we're dealing with math here.

As for where they will meet, well, they started 10 m apart, so if Jane is moving towards John with an acceleration of 0.92 m/s^2 and John is moving towards Jane with an acceleration of 0.767 m/s^2, they will meet closer to Jane. It's like a romantic rendezvous where the distances are reduced as the love accelerates! Ah, young love... or in this case, middle-aged love.

Answer 2

To solve this problem, we need to use Newton's second law of motion and kinematic equations.

(a) To find John's acceleration, we'll first determine the net force acting on Jane.

Using Newton's second law of motion, the net force (F_net) on an object is equal to the product of its mass (m) and acceleration (a), i.e., F_net = m * a.

For Jane:
F_net = m * a
F_net = 50 kg * 0.92 m/s^2
F_net = 46 N

Now, since John is pulling on the rope, the force exerted by John on Jane (F_pull) is equal to the net force acting on Jane.

For John:
F_pull = F_net
F_pull = 46 N

The force exerted by John on Jane is also equal to the mass of John (60 kg) multiplied by his acceleration (a_john).

F_pull = m_john * a_john
46 N = 60 kg * a_john

To find John's acceleration (a_john):
a_john = F_pull / m_john
a_john = 46 N / 60 kg
a_john = 0.77 m/s^2

Therefore, John's acceleration is 0.77 m/s^2.

(b) To find where Jane and John will meet, we can use the equations of motion and substitute the values we have.

One of the equations of motion is:

s = ut + (1/2) a t^2

Where:
s is the displacement
u is the initial velocity
t is the time
a is the acceleration

Since both Jane and John start from rest (initial velocity, u = 0), we can rewrite the equation as:

s = (1/2) a t^2

For Jane:
Jane's displacement, s_jane = 10 m (since they are standing 10 m apart)
Jane's acceleration, a_jane = 0.92 m/s^2

10 = (1/2) * 0.92 * t^2

Simplifying this equation, we can solve for the time (t):

t^2 = (2 * 10) / 0.92
t^2 = 21.74
t ≈ √21.74
t ≈ 4.66 s

Now, we can find the distance covered by John using the equation of motion:

s_john = u_john * t + (1/2) * a_john * t^2

Since John's initial velocity (u_john) is also 0, we have:

s_john = (1/2) * a_john * t^2
s_john = (1/2) * 0.77 * (4.66)^2
s_john ≈ 4.54 m

Jane and John will meet approximately 4.54 m from John's starting position.

Answer 3

A.)

F=ma
F= (50kg)(.92m/s^2)
F= 46N

a= 46N/60kg
a= 0.77m/s^2 toward Jane