A 12 kg bag of hydro spanners starts with a velocity of 6 m/s and moves in a straight line asdescribed by€v = 6 + 3t + 2t 3 . What is the net force on the bag at 4 seconds? How much work isdone on the bag in the first 3 seconds?

To find the net force on the bag at 4 seconds, we need to calculate the derivative of the velocity function with respect to time.

Given that the velocity function is v = 6 + 3t + 2t^3, we can differentiate it to find the acceleration function.

Taking the derivative of v with respect to t, we get a = 3 + 6t^2.

To find the net force, we can use Newton's second law, which states that force (F) is equal to mass (m) multiplied by acceleration (a).

Given that the mass of the bag is 12 kg, we can substitute the values into the equation:

F = m * a F = 12 kg * (3 + 6(4)^2) F = 12 kg * (3 + 6 * 16) F = 12 kg * (3 + 96) F = 12 kg * 99 F = 1188 N

Therefore, the net force on the bag at 4 seconds is 1188 N.

To calculate the work done on the bag in the first 3 seconds, we need to integrate the velocity function with respect to time over the interval from 0 to 3 seconds.

The work done (W) is given by the equation:

W = ∫(v) dt

Substituting the velocity function v = 6 + 3t + 2t^3, we can integrate it over the interval from 0 to 3 seconds:

W = ∫(6 + 3t + 2t^3) dt

Integrating term by term, we get:

W = 6t + (3/2)t^2 + (1/2)t^4

Evaluating the integral from 0 to 3 seconds, we have:

W = (6 * 3) + (3/2 * 3^2) + (1/2 * 3^4) W = 18 + (3/2 * 9) + (1/2 * 81) W = 18 + 13.5 + 40.5 W = 72

Therefore, the work done on the bag in the first 3 seconds is 72 Joules.

To find the net force on the bag at 4 seconds, we need to calculate the derivative of the velocity function with respect to time.

Given that the velocity function is v = 6 + 3t + 2t^3, we can differentiate it to find the acceleration function.

Taking the derivative of v with respect to t, we get a = 3 + 6t^2.

To find the net force, we can use Newton's second law, which states that force (F) is equal to mass (m) multiplied by acceleration (a).

Given that the mass of the bag is 12 kg, we can substitute the values into the equation:

F = m * a F = 12 kg * (3 + 6(4)^2) F = 12 kg * (3 + 6 * 16) F = 12 kg * (3 + 96) F = 12 kg * 99 F = 1188 N

Therefore, the net force on the bag at 4 seconds is 1188 N.

To calculate the work done on the bag in the first 3 seconds, we need to find the area under the velocity-time graph.

The work done (W) is equal to the integral of the force (F) with respect to time (t) over the given time interval.

Given that the force function is F = 6 + 3t + 2t^3, we can integrate it over the interval from 0 to 3 seconds:

W = ∫(0 to 3) (6 + 3t + 2t^3) dt

Evaluating the integral, we get:

W = [6t + (3/2)t^2 + (1/2)t^4] from 0 to 3

W = (6(3) + (3/2)(3)^2 + (1/2)(3)^4) - (6(0) + (3/2)(0)^2 + (1/2)(0)^4)

W = (18 + (3/2)(9) + (1/2)(81)) - (0 + 0 + 0)

W = 18 + (27/2) + (81/2)

W = 18 + 13.5 + 40.5

W = 72

Therefore, the work done on the bag in the first 3 seconds is 72 Joules.