Question

A constant force on an object of mass 3.2 kg for duration of 6 sec.if the velocity changes from 2.5m/s to 9m/sfind the magnitude of the force.

Answer 1

$\bf{\underline{\underline \green{Solution:-}}}$

$\frak\red{Given}\begin{cases}\sf\red{\underline{M = 3.2\:Kg}} \\ \sf\blue{\underline{T = 6s}} \\\sf\green{\underline{U = 2.5 m/s}}\\\sf\orange{\underline {v = 9 m/s}}\end{cases}$

$\underline\mathtt \orange{Formula\:used\:here:-}$

$\bigstar \: \boxed{ \sf \: Force = \frac{m(v - u)}{t } }$

$\underline\mathtt \orange{Putting \: the \: values :-}$

$\longrightarrow \sf {Force = 3.2\: ( 9 - 2.5 )/6 }$

$\longrightarrow \sf {Force = 3.46 N}$

$\underline\mathtt \orange{Therefore:}$

• A force of 3.46 N is exterted on the body.
• The magnitude of force is 3.46 N.

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\thicklines\put(1,1){\line(1,0){6.5}}\put(1,1.1){\line(1,0){6.5}}\end{picture}$