Help Jane & John Pull a Rope: Solve Acceleration & Meetup

[fitchguy316]

Here is a question:

Jane and John with masses of 51kg and 62kg respectively, stand on a frictionless surface 13m apart. John pulls on a rope that connects him to JAne, giving jane an acceleration of .92 m/s*s toward him. a. What is Johns acceleration?? b. IF the pulling force is applied constantly, where will Jane and John meet?

a. f=ma, manipulating this you get John's Acceleration at .756m/s*s

b. More towards John. I got 5.9m away from him but I believe this is wrong..Please help ASAP

 

[rl.bhat]

Both are starting from rest simultaneously.When they meet they must have traveled for the same time intrerval. Use the equation x = 1/2*a*t^2 for both and find x

 

[ogb p]

a. Using the formula F=ma, we can calculate John's acceleration as follows:

F = (51kg + 62kg)(0.92 m/s^2) = 108.74 N

a = F/m = 108.74 N / 62 kg = 1.75 m/s^2

b. To determine where Jane and John will meet, we can use the formula for displacement, s = ut + 1/2at^2, where u is the initial velocity, a is the acceleration, and t is the time.

Since Jane starts at rest and accelerates towards John, her initial velocity is 0 m/s. We can rearrange the formula to solve for t:

t = sqrt(2s/a)

Substituting in the values, we get:

t = sqrt(2(13m)/(0.92 m/s^2)) = 4.26 seconds

Therefore, Jane and John will meet after 4.26 seconds. To find the distance from John, we can use the formula s = ut + 1/2at^2 again:

s = (1.75 m/s^2)(4.26 s)^2 = 31.1 m

So, Jane and John will meet 31.1 meters away from John's initial position.

I hope this helps!