Work Done on 5.8kg Particle in 4.4 seconds

[pringless]

A time-varying net force acting on a 5.8kg particle causes the object to have a displacement given by

x = a + bt + dt^2 + et^3

where a = 2m, b = 1.8 m/s, d = -1.6 m/s^2, e = .94 m/s^3 with x in meteres and t in seconds. Find the work done on the partciel in the first 4.4 seconds of motion. Answer in units of J.

i've taken both derivative and integral, I am not getting any way correct grr. can someone help

 

[sridhar_n]

...

Heres how u do the problem...


\int dw = \int_{x_{1}}^{x_{2}} F.dx; where F is the force and is equal to m*a, where m is the mass and a is the accln.
To find the work done by the force in 1st 4.4 seconds you will have to do this:

x = a + bt + dt^2 + et^3. Therefore:

dx = (b + 2dt + 3et^2)dt

substitute this in the work equation to get:

\int dw = \int_{0}^{4.4} F * (b + 2dt + 3et^2)dt

where:
F = m*a, and a = d^2x for the 1st 4.4 seconds. Find this force and then substitute in the work integral..


There is another way to this problem:
work done = change in kinetic energy. Find the kinetic energy at t = 0 and at t = 4.4 by finding the velocities at t = 0 and at t = 4.4sec by differentiating the expression for x and then substituting for t.

work done = W = m(v_{4.4}^2 - v_{0}^2)/2...

Sridhar

 

[M98Ranger]

To find the work done on the particle, we can use the work-energy theorem which states that the work done on an object is equal to the change in its kinetic energy.

In this case, the kinetic energy of the particle can be calculated using the formula:

KE = 1/2 * m * v^2

where m is the mass of the particle and v is its velocity.

Since the particle has a mass of 5.8kg, we can calculate its kinetic energy at any given time using the above formula. However, since we are interested in the work done in the first 4.4 seconds, we will calculate the kinetic energy at t = 4.4 seconds.

First, we need to calculate the velocity of the particle at t = 4.4 seconds using the given equation for displacement:

x = a + bt + dt^2 + et^3

Substituting t = 4.4 seconds and the given values for a, b, d, and e, we get:

x = 2 + 1.8(4.4) - 1.6(4.4)^2 + 0.94(4.4)^3 = 38.67 meters

Now, to calculate the velocity, we take the derivative of the displacement equation:

v = b + 2dt + 3et^2

Substituting t = 4.4 seconds and the given values for b, d, and e, we get:

v = 1.8 + 2(-1.6)(4.4) + 3(0.94)(4.4)^2 = -11.648 m/s

Finally, we can calculate the kinetic energy at t = 4.4 seconds:

KE = 1/2 * m * v^2 = 1/2 * 5.8 * (-11.648)^2 = 380.16 J

Therefore, the work done on the particle in the first 4.4 seconds of motion is 380.16 Joules.