Acceleration as a Function of Distance

[bollocks748]

Homework Statement

Consider a block of mass M that is pulled up an incline by a force T that is parallel to the surface of the incline. The block starts from rest and is pulled a distance x by the force T. The incline, which is frictionless, makes an angle theta with respect to the horizontal.

1.1 Write down the work done by the force T.


1.2 Calculate the potential energy of the block as a function of the position x. From the work and the potential energy calculate the kinetic energy of the block as a function of position x.


1.3 Calculate the acceleration, a, of the block as a function of position x. Calculate the velocity of the block from the acceleration and hence the kinetic energy as a function of position. Show that the kinetic energies calculated in these two ways are equivalent.


1.4 Calculate the ratio of the potential to kinetic energy. What happens to this ratio for large T? Check your answer for the case theta = pi/2 which you should be able to recalculate easily.

Homework Equations

I didn't have problems with the first two parts, but I might as well put them here just in case:

Work done by T = T*x

PE(x)= m*g*x*sin(theta)

The Attempt at a Solution

What I'm really having trouble with is defining acceleration as a function of x, in part 3. I've had to do it before using the chain rule, but most of the time I was given a function of v(x), and integrated it, or used dv/dx and dx/dt to find dv/dt, or something of that nature. But I've never encountered a problem where I wasn't given some starter formula to work with... and I'm really at a loss here. Please help!

 

[Hootenanny]

What I'm really having trouble with is defining acceleration as a function of x, in part 3. I've had to do it before using the chain rule, but most of the time I was given a function of v(x), and integrated it, or used dv/dx and dx/dt to find dv/dt, or something of that nature. But I've never encountered a problem where I wasn't given some starter formula to work with... and I'm really at a loss here. Please help!

You're on the right lines with use of the chain rule. Note that although you haven't been given an equation explicitly, you can formulate one yourself. Consider applying Newton's Second Law to the block.

Be aware that T is not the only force acting on the block.

 

[baba89]

Acceleration is defined as the rate of change of velocity with respect to time. In this problem, we are looking for the acceleration as a function of distance, which means we need to find the rate of change of velocity with respect to distance.

To do this, we can use the formula for acceleration, a=dv/dt, and the chain rule to find the derivative of velocity with respect to distance. Using the chain rule, we get:

a = dv/dt * dt/dx

Since the block is being pulled up the incline at a constant speed, the velocity is also constant, so dv/dt = 0. Therefore, we are left with:

a = 0 * dt/dx = 0

This means that the acceleration of the block is zero as a function of distance. This makes sense, as the block is being pulled up the incline at a constant speed, so its acceleration should be zero.

To find the velocity of the block, we can use the formula v = dx/dt. Since the block is being pulled at a constant speed, the velocity is also constant, so v = dx/dt = constant. This also makes sense, as the block is being pulled at a constant speed, so its velocity should also be constant.

To show that the kinetic energies calculated in parts 2 and 3 are equivalent, we can use the formula for kinetic energy, KE= 1/2*m*v^2. Since the velocity is constant, we can take it out of the equation, and we are left with:

KE = 1/2*m*v^2 = 1/2*m*(dx/dt)^2 = 1/2*m*(dx/dt)^2*dt/dx = 1/2*m*v^2

This shows that the kinetic energy calculated using the acceleration as a function of distance is equivalent to the kinetic energy calculated using the velocity as a function of distance.

For part 1.4, the ratio of potential energy to kinetic energy can be calculated by dividing the potential energy by the kinetic energy. Using the formulas from parts 1.2 and 1.3, we get:

PE/KE = (m*g*x*sin(theta))/(1/2*m*v^2) = 2*g*x*sin(theta)/v^2

As force T increases, the velocity of the block will also increase, making the ratio of potential to kinetic energy decrease. This makes