Answer to Question #86542 in Mechanics | Relativity for Udeh George

Question

A block of weight 7 newton rests on a level floor. the frictional force between the block and the floor is 1 newton. a horizontal force of 1.4 newton is used to pull the block for 4 seconds. what is the velocity of the block after this time?

Answer 1

Given: 

$P = 7~\text{N};~F_{fric} = 1~\text{N};~F_{pull}= 1.4~\text{N};~t = 4~\text{s}. \\$

According to the 2nd Newton's law,

$ma = F = F_{pull} - F_{fric}; \\$

To express the mass of the block via the weight, 

$P = mg; \implies m = \frac{P}{g}; \\$

Velocity of the block, assuming the block was initially at rest (initial velocity zero), 

$v = at = \frac{F}{m}t = \frac{F}{P}gt = \frac{F_{pull} - F_{fric}}{P}gt = \frac{1.4~\text{N}~-~1~\text{N}}{7~\text{N}}~*~9.8~\frac{\text{m}}{\text{s}^2}~*~4~\text{s} = 2.24~\frac{\text{m}}{\text{s}}.$