Question

A constant force of 4.5N acts on a 7.2 kg object for 10 s what is the change in the object velocity

Answer 1

The change in the object's velocity is 6.25 m/s

Step-by-step explanation:

Force

According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:

F = ma [1]

Where a is the acceleration of the object.

On the other hand, the equations of the Kinematics describe the motion of the object by the equation:

`v_f=v_o+at\qquad\qquad[2]`

Where:

vf is the final speed

vo is the initial speed

a is the acceleration

t is the time

It's given a constant force of F=4.5 N acts on an object of mass m=7.2 kg. The object has an acceleration we can calculate solving [1] for a:

`a=(F)/(m)`

`a=(4.5)/(7.2)=0.625`

`a=0.625\ m/s^2`

It's required con compute the change of velocity (or its magnitude, the speed). From [2]:

`v_f-v_o=at`

The expression at the left side is the change of speed Δv:

`\Delta v=at`

Knowing the time is t=10 s:

`\Delta v=0.625*10`

`\Delta v=6.25\ m/s`

The change in the object's velocity is 6.25 m/s