Question

1. a 2.0 newton net external force is applied to an object for three seconds. after that, the applied net external force decreases to 0.0 newtons from three to five seconds. assume the object has a mass of 5.0 kilograms.

a. what is the impulse (change in momentum) of the object from 0 to 5 seconds.?
b. what is the objects change in velocity from 0 to 5 seconds?
c. what is the acceleration of the object for the first three seconds?

Explanation:

Part a

Step1: Calculate impulse for first 3 seconds

Impulse \(J = F\times t\). For \(F = 2\space N\) and \(t=3\space s\), \(J_1=2\times3 = 6\space N\cdot s\)

Step2: Calculate impulse for 3 - 5 seconds

The area of the triangle from \(t = 3\space s\) to \(t=5\space s\). The formula for the area of a triangle is \(A=\frac{1}{2}\times base\times height\). Here, base \(b=(5 - 3)=2\space s\) and height \(h = 2\space N\). So \(J_2=\frac{1}{2}\times2\times2= 2\space N\cdot s\)

Step3: Calculate total impulse

Total impulse \(J=J_1 + J_2\). Substitute \(J_1 = 6\space N\cdot s\) and \(J_2=2\space N\cdot s\). So \(J=6 + 2=8\space N\cdot s\)

Step1: Use impulse - momentum theorem

The impulse - momentum theorem is \(J=\Delta p=m\Delta v\). We know \(J = 8\space N\cdot s\) and \(m = 5\space kg\). Rearranging for \(\Delta v\), we get \(\Delta v=\frac{J}{m}\)

Step2: Calculate \(\Delta v\)

Substitute \(J = 8\space N\cdot s\) and \(m = 5\space kg\) into \(\Delta v=\frac{J}{m}\). So \(\Delta v=\frac{8}{5}=1.6\space m/s\)

Step1: Use Newton's second law

Newton's second law is \(F = ma\). Rearranging for \(a\), we get \(a=\frac{F}{m}\)

Step2: Calculate acceleration

For \(F = 2\space N\) and \(m = 5\space kg\), \(a=\frac{2}{5}=0.4\space m/s^{2}\)

Answer:

\(8\space N\cdot s\)

Part b