Question

a constant force acts on an object of mass 10kg for a duration of 5 second. it increases the object velocity from 6m/sec to 14m/sec. find the magnitude of the applied force​

Answer 1

Answer:

• 16 N

Given :

• Mass (m) = 10kg
• Time (t) = 5 s
• Initial velocity (u) = 6 m/s
• Final velocity (v) = 14 m/s

To find:

• Magnitude of force (F)

Solution:

Change in velocity = Final velocity - Initial velocity

= 14 - 6

= 8 m/s

Acceleration (a) is given by :

$\tt {acceleration}\: = \dfrac{change \: in \: velocity}{time}$

$\tt \: a = \dfrac{8}{5} m {s}^{ - 2}$

The magnitude of force is given by the formula :

$\boxed{\tt \purple{force = mass \times acceleration}}$

$\tt = \cancel{10} \times \frac{8}{ \cancel{5}} \\ \\ \tt = 2 \times 8$

= 16 N

$\underline{\sf \therefore The\ magnitude\ of\ force\ is\ 16\ N}$

_________________________

• A push or pull is called a force . It brings about a change in position , direction and shape of the body .

• It is defined as the product of mass and acceleration of the body .

• Its SI unit is Newton denoted by N .

• Its CGS unit is dyne .