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sleet_krkn [62]
3 years ago
9

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 frict

ion 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?
Physics
1 answer:
Firdavs [7]3 years ago
7 0

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

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A small fish is dropped by a pelican that is rising steadily at 0.50 m/s.
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(b)  31 m

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From Newton's equation of motion,

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v = u + gt ................... Equation 1

Where v = final velocity, u = initial velocity, g = acceleration due to gravity, t = time.

Note: Let upward velocity be negative and downward be positive

Given: u = -0.5 m/s (upward), t = 2.5 s

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Substitute into equation 1

v = 0.5+9.81(2.5)

v = -0.5+24.525

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(b) using

s₁ = ut + 1/2gt²......................... Equation 2

Where s₁ = distance at which the fish fall after being dropped by the pelican

Given: u = - 0.5 m/s, t = 2.5 s, g = 9.81 m/s²

Substitute into equation 2

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s₂ = vt ................ Equation 3

Where s₂ = the distance by which the pelican rise during this time.

Given: v = 0.5 m/s, t= 2.5 s

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s₂ = 1.25 m.

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Distance between the pelican and fish  = 29.41+1.25

Distance between the pelican and fish  = 30.66

Distance between the pelican and fish ≈ 31 m

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