Question

1. refer to the force versus time graph of an object with a mass of 14.1 kg.

a. what is the impulse (change in momentum) of the object from 0 to 8 seconds?
b. what is the objects change in velocity from 5 to 8 seconds?
c. what is the objects acceleration from 5 to 8 seconds?

Explanation:

Step1: Calculate impulse from 0 - 8s (area under F - t graph)

The area under the force - time graph is composed of a triangle (base \(b = 2s\), height \(h=4N\)) and a trapezoid (parallel sides \(a = 4N\), \(b = 1N\), height \(h=(5 - 2)=3s\)) and a rectangle (length \(l=(8 - 5)=3s\), width \(w = 1N\)).
The area of the triangle \(A_1=\frac{1}{2}\times2\times4 = 4N\cdot s\).
The area of the trapezoid \(A_2=\frac{(4 + 1)}{2}\times3=\frac{5\times3}{2}=7.5N\cdot s\).
The area of the rectangle \(A_3=1\times3 = 3N\cdot s\).
The total impulse \(J=A_1+A_2+A_3\).
\(J=4 + 7.5+3=14.5N\cdot s\).

Step2: Calculate impulse from 5 - 8s (area of rectangle)

For the time interval \(5 - 8s\), the force \(F = 1N\) and time interval \(\Delta t=8 - 5=3s\).
The impulse \(J=\int_{t_1}^{t_2}Fdt=F\Delta t\) (since \(F\) is constant). So \(J = 1\times3=3N\cdot s\).
Using the impulse - momentum theorem \(J=\Delta p=m\Delta v\), where \(m = 14.1kg\).
\(\Delta v=\frac{J}{m}=\frac{3}{14.1}\approx0.213m/s\).

Step3: Calculate acceleration from 5 - 8s

Using Newton's second law \(F = ma\) (since \(F\) is constant from \(5 - 8s\), \(F = 1N\)).
\(a=\frac{F}{m}\), substituting \(m = 14.1kg\) and \(F = 1N\).
\(a=\frac{1}{14.1}\approx0.0709m/s^{2}\).

Answer:

a. The impulse (change in momentum) from \(0\) to \(8\) seconds is \(14.5N\cdot s\).
b. The object’s change in velocity from \(5\) to \(8\) seconds is approximately \(0.213m/s\).
c. The object’s acceleration from \(5\) to \(8\) seconds is approximately \(0.0709m/s^{2}\).