Question

A0.25 kg puck is initially stationary on an ice surface with negligible friction. At time t=0, a horizontal force begins to move the puck. The force is given by F = (12-3.00) with F in newtons and t in seconds, and it acts until its magnitude is zero. (a) What is the magnitude of the impulse on the puck from the force between t=0.500 s and t=1.25 s ?

Answer 1

The magnitude of the impulse on the puck between t=0.500 s and t=1.25 s is calculated using the integral of the force function and is found to be 7.03125 kg·m/s.

Step-by-step explanation:

To calculate the magnitude of the impulse on the puck from t=0.500 s to t=1.25 s, we use the equation for impulse, which states that impulse is the change in momentum, or the integral of force over time. The force acting on the puck is given as F = (12 - 3t) N.

We need to integrate this force from t=0.500 s to t=1.25 s:

Impulse (J) = ∫0.5001.25F dt = ∫0.5001.25(12 - 3t) dt

Calculating the definite integral yields:

J = [12t - (3/2)t2]0.5001.25

J = [12(1.25) - (3/2)(1.25)2] - [12(0.500) - (3/2)(0.500)2]

J = (15 - 2.34375) - (6 - 0.375)

J = 12.65625 - 5.625

J = 7.03125 kg·m/s

Therefore, the magnitude of the impulse experienced by the puck from t=0.500 s to t=1.25 s is 7.03125 kg·m/s.