Respuesta :
Answer:
t = 23.92 s
Explanation:
Newton's second law:
∑F = m*a Formula (1)
∑F : algebraic sum of the forces in Newton (N)
m : mass s (kg)
a : acceleration (m/s²)
We define the x-axis in the direction parallel to the movement of the block and the y-axis in the direction perpendicular to it.
Forces acting on the block
W: Weight of the block : In vertical direction, downward
FN : Normal force : perpendicular to the floor, upward
fk : Kinetic friction force: parallel to the floor and opposite to the movement
F = 86.4 N , in the direction of the motion
Calculated of the W
W= m*g
W= 16.8 kg* 9.8 m/s² = 164.64 N
Calculated of the FN
We apply the formula (1)
∑Fy = m*ay ay = 0
FN - W = 0
FN = W
FN = 164.64 N
Calculated of the fk
fk = μk*FN
fk = 0.426* 164.64 N
fk = 70.13 N
We apply the formula (1) to calculated acceleration of the block:
∑Fx = m*ax , ax= a : acceleration of the block
F - fk = m*a
86.4 -70.13 = (16.8)*(-a)
16.26 = (16.8)*(-a)
a = -(16.26 )/ (16.8)
a = - 0.97 m/s²
Kinematics of the block
Because the block moves with uniformly accelerated movement we apply the following formula :
vf = v₀ + a*t Formula (2)
Where:
t: time interval (m)
v₀: initial speed (m/s)
vf: final speed (m/s)
Data:
v₀ = 23.2 m/s
vf = 0
a = -0.97 m/s²
Time it takes for the block to stop
We replace data in the formula (2) to calculate the time
vf= v₀+a*t
0 = 23.2+( -0.97)*t
(0.97)*t = 23.2
(0.97)*t = 23.2
t = 23.2 / (0.97)
t = 23.92 s
The equation of motion is the relation between the distance, velocity, acceleration and time of a moving body.
The time taken for the block to come to stop is 23.92 seconds.
What is the third equation of motion?
The equation of motion is the relation between the distance, velocity, acceleration and time of a moving body.
The third equation of the motion for velocity can be given as,
[tex]v=u+at[/tex]
Here, [tex]u[/tex] is the initial body. [tex]a[/tex] is the acceleration of the body and [tex]t[/tex] is the time taken by it.
Given information-
The mass of the block is 16.8 kg.
The initial velocity of the block is 23.2 m/s.
The coefficient of kinetic friction (μk) is 0.426.
The magnitude of the force is 86.4 N.
The normal force on the block is equal to the weight of the block thus, the normal force is,
[tex]F_N=16.8\times9.8\\F_N=164.64 \rm N[/tex]
As the kinetic friction force is the product of the normal force and coefficient of kinetic friction. Thus,
[tex]F_k=164.64\times0.462\\F_k=70.13 \rm N[/tex]
To find the acceleration,
[tex]F-f_k=ma[/tex]
Put the values,
[tex]86.4-70.13=16.8\times a\\a=-0.97\rm m/s^2[/tex]
The final velocity using the third law of equation can be given as,
[tex]v_f=u+at[/tex]
Put the values,
[tex]0=23.2+(-0.97)t\\t=23.92[/tex]
Hence, the time taken for the block to come to stop is 23.92 seconds.
Learn more about the equation of motion here;
https://brainly.com/question/13763238
