Need help with determining acceleration when given a distance

[JDM5000]

Homework Statement

A machine has a steel shuttle with mass m that is pulled along a vertically positioned square steel rail by a spring with a spring constant k. The shuttle is released when the spring acts on the shuttle with force equal to F at a 45 degree angle.
a) what is the initial acceleration of the shuttle?
b) what will be the acceleration of the shuttle after it moved a distance L?

Homework Equations

F=ma ; X=vt+(1/2)at^2 ; u_k(n)=f_k ;

The Attempt at a Solution

For A
Ʃx= K*cos(45)-f_k=m_x*a
Ʃy=K*sin(45)+N-mg=0

a= (kcos45-uk(mg-Ksin45))/m
a= (t/m)(kcos45-ksin45)(uk*g)

For B)
Wouldn't it just be my a_1 subtracted by the new value of a?
I was thinking since theta is going to change with the elastic band.

 

[anathema]

So you can find the new x and y components of the band. Once you find those you can find the new a_2.

Hello, thank you for your forum post. Let me provide some insight on your questions.

a) To find the initial acceleration of the shuttle, we can use the formula F=ma, where F is the force acting on the shuttle, m is the mass of the shuttle, and a is the acceleration. In this case, the force acting on the shuttle is the force from the spring, which is given as F at a 45 degree angle. We can use vector decomposition to find the x and y components of this force, which will be Fcos45 and Fsin45 respectively. Therefore, the equation becomes:

Fcos45=ma

Solving for a, we get:

a= Fcos45/m

b) To find the acceleration of the shuttle after it has moved a distance L, we can use the formula X=vt+(1/2)at^2, where X is the distance travelled, v is the initial velocity, a is the acceleration, and t is the time. In this case, we can assume that the initial velocity is 0, as the shuttle is released from rest. Therefore, the equation becomes:

X=(1/2)at^2

We know that the distance travelled is L, so we can substitute that in the equation:

L=(1/2)at^2

Solving for a, we get:

a= (2L)/t^2

Now, we need to find the time t. We can use the formula t=√(2L/a), where L is the distance travelled and a is the acceleration. Substituting the values, we get:

t=√(2L/((Fcos45/m))

Therefore, the final equation for the acceleration after the shuttle has moved a distance L is:

a= (Fcos45/m)/(√(2L/((Fcos45/m)))

I hope this helps in solving your problem. Good luck!