A block of mass 200 kg is initially at rest at x0 on a...

A block of mass 2.00 kg is initially at rest at x=0 on a slippery horizontal surface for which there is no friction. Starting at time t=0, a horizontal force Fx(t) = β - αt is applied to the block, where α = 6.00 N/s and β = 4.00 N. 1- How long from t = 0 does it take the block to return to x = 0, and what is its speed at this point? 2- What is its speed at this point?

Transcript

00:01 The mass of the block is 2kg.

00:03 The force acting on the given block is varying with respect to time.

00:07 It is a function of time and f of t is equal to beta minus alpha t.

00:13 Now the block is returning to its initial position after some time and we have to find out the value such that beta is equal to 4 newton and alpha is equal to 6 newton per second.

00:30 So first of all, let us try to find out what is the velocity, what is the displacement such that we can define at what time it is going to return to its initial position.

00:40 So to find out that, we know that by newton's second law of motion, f is equal to mass time's acceleration, but acceleration is nothing but rate of change of velocity.

00:58 So we can write it as mdb by d t.

01:00 But according to the given question, forces a function of time and it can be written.

01:04 As beta minus alpha t so beta subtracted with alpha t is equal to m dv by d t or dv is nothing but one divided by m times beta minus alpha t d t now we can integrate on both the sides so that we get the velocity function integrating from initial velocity u to final velocity v dv is equal to one divided by m m is a constant so we can keep it out of integration zero to t beta minus alpha t d t so the velocity function is going to be v minus u 1 divided by m integration of beta d t is going to be beta t minus alpha integration of t d t is t squared divided by 2 on substituting the limit 0 to t we are going to get this function now initially the block is at rest so u value is equal to 0 so we get v value v is equal to 1 divided by 2 beta value is 4 minus alpha is 6 so this is going to be equal to v is equal to 1 divided by 2 40 minus 3 t squared so this is going to be our velocity now this is the velocity function now let us find out the displacement function as well for that let us call this equation as 1 so to find out the displacement further we know that displacement is nothing but velocity is nothing but rate of change of displacement.

02:43 So v is equal to ds by dt.

02:46 So again, d s can be written as v, dt.

02:50 So this is going to be now equal to ds, again integrating from 0 to s.

02:57 That is equal to velocity which is half times 40 minus 3t squared dt integrating from 0 to t.

03:10 So on integration displacement is going to be equal to s and half is constant...

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