Question

A stationary 1.67-kg object is struck by a stick. The object experiences a horizontal force given by F = at - bt^2, where t is the time in seconds from the instant the stick first contacts the object. If a = 15 N/s and b = 20 N/s^2, what is the impulse of the force over the period 0 s ≤ t ≤ 3.0 s?

a) 54 Ns
b) 90 Ns
c) 135 Ns
d) 180 Ns

Answer 1

The impulse of the force over the given time interval is 90 Ns.

Step-by-step explanation:

To find the impulse of the force, we need to calculate the integral of the force function over the given time interval. The formula for impulse is:

Impulse = ∫(F dt)

Plugging in the values for a and b, we get:

F = 15t - 20t^2

Integrating this equation from 0 s to 3.0 s:

Impulse = ∫(15t - 20t^2) dt

Evaluating the integral gives us an impulse of 90 Ns.