A single force acts on a 5.0 kg particle-like object in such a way that the position of the object as a function of time is given by x = 3.0t - 4.0t 2 + 1.0t 3, with x in meters and t in seconds. Find the work done on the object by the force from t = 0 to t = 2.0 s.

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* Physics/Math - bobpursley, Thursday, March 1, 2007 at 6:28am

find the position at at t=2 and t=0. The difference in those is the displacement.

Work= force times displacement.

* Physics/Math - Technoboi11, Friday, March 2, 2007 at 12:11am

So when t=0, x=0.
When t=2, x=-2
displacement= -2-0 = -2m

W=Fd
F=ma=(5)(9.8)=49

W=(49 N)(-2 m)
W= -98 J

This is the answer that I got but it is an incorrect answer. Is there something wrong with my calculations???

x= 3.0t - 4.0t^ 2 + 1.0t^3
dx/dt= 3 - 8 t + 3t^2

d"x/dt"= acceleration= -8 + 6 t

F= ma = m (-8+6t)
INT F dx= work
work = INT m (-8+6t)(3 - 8 t + 3t^2)dt

integrate from t=0 to t= 2.

check my thinking.

Can you use W=1/2mv1^2-1/2mv2^2
Use 3-8t+3t^2 and solve for v1 when t=0 and v2 when t=2. Then plug v1 and v2 into the above equation with 5 as mass?

I used the above equation and it worked

Use your last equation. Do not use d"x/dt" because this force is a variable force, and F = ma is not applicable here, I think.

To find the work done on the object by the force from t = 0 to t = 2.0s, you can use the formula:

Work = force x displacement.

Now, let's start by finding the displacement of the object. The given equation of the position as a function of time is:

x = 3.0t - 4.0t^2 + 1.0t^3.

To find the displacement, we need to find the position at t = 2s and t = 0s:

When t = 0, x = 3.0(0) - 4.0(0)^2 + 1.0(0)^3 = 0.
When t = 2, x = 3.0(2) - 4.0(2)^2 + 1.0(2)^3 = 12 - 16 + 8 = 4.

The displacement is the difference between the final position and the initial position, so the displacement is:

displacement = 4 - 0 = 4m.

Next, we need to find the force. The force is given by Newton's second law, F = ma, where m is the mass of the object and a is the acceleration.

Given that the object is particle-like and has a mass of 5.0kg, the force can be calculated as:

F = ma = 5.0kg * (acceleration).

To find the acceleration, we differentiate the equation of the position with respect to time (dx/dt):

dx/dt = 3 - 8t + 3t^2.

Now, taking the second derivative (d^2x/dt^2) gives us the acceleration:

d^2x/dt^2 = -8 + 6t.

Substituting this into the equation for force, we have:

F = 5.0kg * (-8 + 6t).

Now, we can calculate the work by multiplying the force by the displacement:

Work = F * displacement
= (5.0kg * (-8 + 6t)) * 4m.

To find the work from t = 0 to t = 2s, we need to integrate the force with respect to displacement over the given time interval:

Work = ∫[(5.0kg * (-8 + 6t))] dt, integrated from t = 0 to t = 2.

Evaluating this integral will give us the value of work done on the object by the force over the given time interval.

As an alternative approach, you mentioned using the equation W = 1/2mv1^2 - 1/2mv2^2, where v1 and v2 are the velocities at times t = 0 and t = 2, respectively. To use this equation, you need to find the velocities by taking the derivative of the position equation with respect to time.

After finding v1 and v2, you can substitute them into the equation and solve for work, using the mass value of 5.0kg.

Both of these methods should give you the correct answer. Please check your calculations and make sure you're evaluating the integrals or the equation values correctly to obtain the correct work value.