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
