At t = 0, an object of mass m is at rest at x = 0 on a...

Name: At t = 0, an object of mass m is at rest at x = 0 on a horizontal, frictionless surface. A horizontal force Fx = F0(1 - 2t/T), which decreases from t = 0 to t = T, is exerted on the object. Find an expression for the object's velocity and position at time T.
Uploaded: 2022-06-07T19:31:56-08:00
Duration: 4 min 33 s
Channel: Lucas Finney
Description: At t = 0, an object of mass m is at rest at x = 0 on a horizontal, frictionless surface. A horizontal force Fx = F0(1 - 2t/T), which decreases from t = 0 to t = T, is exerted on the object. Find an expression for the object's velocity and position at time T.

At t = 0, an object of mass m is at rest at x = 0 on a horizontal, frictionless surface. A horizontal force Fx = F0(1 - 2t/T), which decreases from t = 0 to t = T, is exerted on the object. Find an expression for the object's velocity and position at time T.

Transcript

00:01 For this problem, we are told that at t equal 0, an object of mass m is at rest at x equal 0 on a horizontal frictionless surface.

00:08 A horizontal force, fx equals f0 times 1 minus 2t over big t, which decreases from t equals 0 to t equals big t, is exerted on the object, whereas to find an expression for the object's velocity and position at time big t.

00:23 So to begin, we have newton's second law, f equals m .a, which in turn means that a, is equal to f over m now the reason why this matters is we is because we have that velocity or actually i'll say it this way we know that acceleration know that acceleration is the time derivative of velocity which means that velocity must be equal to the integral of the acceleration in this case we'll be integrating from little from zero up to big t and we'd be integrating well the acceleration is going to be f of t over m we're integrating that over t so this is going to be the integral from zero to big t of f not so this we can write this as f not minus two over big t or minus two f not over big t times little t d t so integrating f not well that's just a constant we'd get f not times little t then we'd have minus two f not over big t all of those are all constants and then integral of little t is going to become t squared over two so that just divides out with that two out front and that is going to be evaluated from zero up to big t actually one moment here i'm going to note i'll back up for a second here so since that is the result of finding our anti -derivative can say that v of t is equal to f -not little t minus minus f -not t squared over big t, where we don't need to concern ourselves with the constant because we know that v of t should be equal to zero at zero, so this is satisfying that.

02:28 That being said, we can find v at big t just by plugging in our value for t...

Question

At t = 0, an object of mass m is at rest at x = 0 on a horizontal, frictionless surface. A horizontal force Fx = F0(1 - 2t/T), which decreases from t = 0 to t = T, is exerted on the object. Find an expression for the object's velocity and position at time T.

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