A block of mass m=16.8 kg is sliding on a surface with initial velocity v=23.2 m/s. The block has a coefficient of kinetic friction μk = 0.426 with respect to the surface and is being pushed by a force of magnitude F=86.4 N in the direction of its motion. Due to kinetic friction, the block is slowing down. How long will it take for the block to come to a stop?

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)

dа ta:

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