Question

part b: calculating net force (5 points - 1 point each)
instructions: calculate the net force using ( f = ma ) for each diagram below.
diagram 1: box and arrow diagram showing a 5 kg object accelerating at ( 2 m/s^{2} )
mass = 5 kg, acceleration = ( 2 m/s^{2} )
net force ( (f = ma) ) = ______ n
diagram 2: a car being pushed forward on a road
mass = 10 kg, acceleration = ( 3 m/s^{2} )
net force ( (f = ma) ) = ______ n
diagram 3: a person kicking a soccer ball
mass = 1.5 kg, acceleration = ( 4 m/s^{2} )
net force ( (f = ma) ) = ______ n
diagram 4: a person pushing a shopping cart
mass = 20 kg, acceleration = ( 0.5 m/s^{2} )
net force ( (f = ma) ) = ______ n
diagram 5: a rocket accelerating upward
mass = 12 kg, acceleration = ( 1.5 m/s^{2} )
net force ( (f = ma) ) = ______ n

Explanation:

Step1: Substitute values for Diagram 1

Given \(m = 5\space kg\) and \(a=2\space m/s^{2}\), use \(F = ma\). So \(F=5\times2\).

Step2: Substitute values for Diagram 2

Given \(m = 10\space kg\) and \(a = 3\space m/s^{2}\), use \(F=ma\). So \(F = 10\times3\).

Step3: Substitute values for Diagram 3

Given \(m=1.5\space kg\) and \(a = 4\space m/s^{2}\), use \(F=ma\). So \(F=1.5\times4\).

Step4: Substitute values for Diagram 4

Given \(m = 20\space kg\) and \(a=0.5\space m/s^{2}\), use \(F=ma\). So \(F=20\times0.5\).

Step5: Substitute values for Diagram 5

Given \(m = 12\space kg\) and \(a = 1.5\space m/s^{2}\), use \(F=ma\). So \(F=12\times1.5\).

Answer:

Diagram 1: \(10\space N\)
Diagram 2: \(30\space N\)
Diagram 3: \(6\space N\)
Diagram 4: \(10\space N\)
Diagram 5: \(18\space N\)