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a 5 kg block is pulled across a table by a horizontal force of 40 n with a frictional force 8 n opposing the motion. calculate the acceleration of the object



Answer :

Answer:

a=6.4m/s^2

Explanation:

from basic terms , this problem can be tied to newtons second law of motion,

which states that

The rate of change in motion is directly proportional to the force applied .

a force is that which tends to change a body's state of rest or uniform motion in a straight line.

frictional force is the force tat oppose the motion between an object

acceleration is the cane in velocity per time

mass is the composition of matter in an object.

f[tex]\alpha[/tex] m(v-u)/t

f=km(v-u)/t

k=1

f=m(v-u)/t

(v-u)/t=a

f=ma

the sum of forces on this block will be

F-Fr=ma

notice that the frictional force is in opposition to the horizontal force, hence it is designated as a negative force

40-8=5a

32=5a

a=32/5

a=6.4m/s^2

Given the forces acting on the 5kg block, its acceleration is 6.4m/s²

Given the data in the question;

Mass of block; [tex]m = 5kg[/tex]

Horizontal force Pulling the block; [tex]F = 40N = 40kg.m/s^2[/tex]

Frictional force;  [tex]F_{fric} = 8N = 8kg.m/s^2[/tex]

From Newton's Second Law:

[tex]F = ma[/tex]

Where F is the force, m is the mass of the object and a is the acceleration.

Now its is stated that the box was pulled, meaning there was displacement.

For displacement to occur, the Force must overcome both the frictional force opposing it and the weight of the box.

Hence; F = Frictional force + Weight of box

[tex]F = F_{fric} + ma[/tex]

We substitute our values into the equation and solve for "a"

[tex]40kg.m/s^2 = 8kg.m/s^2 + ( 5kg * a)\\\\40kg.m/s^2 - 8kg.m/s^2 = ( 5kg * a)\\\\ ( 5kg * a) = 32kg.m/s^2\\\\a = \frac{32kg.m/s^2}{5kg} \\\\a = 6.4m/s^2[/tex]

Therefore, given the forces acting on the 5kg block, its acceleration is 6.4m/s²

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