Question

a constant force acts on an object of mass 3.2 kg for a duration of 6 seconds it increases the velocity from the object from 2.5 M per second to 9.0 M per second. Find the magnitude of the force

Answer 1

Hello!

• $\bf{t}$ = 6s 
• $\bf{v}$ = 9 ms⁻¹
• $\bf{u}$ = 2.5 ms⁻¹
• $\bf{m}$ = 3.2 kg

Then,

F = m $\dfrac{\sf (v - u)}{\sf t}$

F = 3.2 $\dfrac{\sf (9 - 2.5)}{\sf 6}$ = $\boxed{\sf 3.46\: N}$

Hence,
Magnitude of the force is $\bf{3.46\:N}$.

Cheers!