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

Answer:

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

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:

$a=\frac{F}{m}$

$a=\frac{4.5}{7.2}=0.625$

$a=0.625\ m/s^2$

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