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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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Answer:

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Explanation:

a) According to the diagram 1 in the attached image:

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Also:

v_{C} =v_{A}+w_{AC}*r_{C/A}\\v_{Ci}=-3j+\left[\begin{array}{ccc}i&j&k\\0&0&w_{AC} \\6.883&-9.829&0\end{array}\right]\\v_{Ci}=-3j+(0+9.829w_{AC} i-(0-6.883w_{AC})j\\v_{Ci}=9.829w_{AC}i+(-3+6.883w_{AC})j

If we comparing both sides of the expression:

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v_{C}=9.829*0.435=4.28in/s

b) According to the diagram 2 in the attached image:

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v_{C}=v_{A}+w_{AC}r_{C/A}\\v_{C}=-3j+\left[\begin{array}{ccc}i&j&k\\0&0&w_{AC}\\7.713&-9.192&0\end{array}\right] \\v_{Ci}=-3j+(9.192w_{AC})i+7.713w_{AC}j\\v_{Ci}=9.192w_{AC}i+(7.713w_{AC}-3)j

Comparing both sides of the expression:

7.713w_{AC}-3=0\\w_{AC}=0.388rad/s\\v_{C}=3.575i

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A crane lifts up two boxes.<br><br> Which free body diagram shows the forces acting on block A?
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Answer:

The 3rd graph

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The problem asks us to find the free body diagram of block A, so we must find all the forces acting on block A.

We have 3 forces acting on block A in total:

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Two identical positive charges exert a repulsive force of 6.2 × 10−9 n when separated by a distance 3.7 × 10−10 m. calculate the
In-s [12.5K]
The electrostatic force between two charges is given by
F=k \frac{q_1 q_2}{r^2}
where
k is the Coulomb's constant
q1 and q2 are the two charges
r is their separation

In this problem, the two charges are identical, so we can call them q1=q2=q, and the formula becomes
F=k \frac{q^2}{r^2}

Since we know the magnitude of the force and the separation between the two charges, we can re-arrange the equation to find the value of each charge:
q= \sqrt{ \frac{Fr^2}{k} } = \sqrt{ \frac{(6.2 \cdot 10^{-9} N)(3.7 \cdot 10^{-10}m)^2}{8.99 \cdot 10^9 Nm^2C^{-2}} }=3.07 \cdot 10^{-19} C
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Calculate the nuclear binding energy per nucleon for 136^Ba if its nuclear mass is 135.905 amu.
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Answer:

1.312 x 10⁻¹² J/nucleon

Explanation:

mass of ¹³⁶Ba = 135.905 amu

¹³⁶Ba contain 56 proton and 80 neutron

mass of proton = 1.00728 amu

mass of neutron = 1.00867 amu

mass of ¹³⁶Ba = 56 x  1.00728 amu + 80 x 1.00867 amu

                      = 137.10128 amu

mass defect = 137.10128 - 135.905

                    = 1.19628 amu

mass defect = 1.19628 x 1.66 x 10⁻²⁷ Kg

                     = 1.9858 x 10⁻²⁷ Kg

speed of light = 3 x 10⁸ m/s

binding energy,

E = mass defect x c²

E = 1.9858 x 10⁻²⁷ x (3 x 10⁸)²

E = 17.87 x 10⁻¹¹ J/atom

now,

binding energy per nucleon =\dfrac{17.87\times 10^{-11}}{136}

                                              = 0.1312 x 10⁻¹¹ J/nucleon

                                              = 1.312 x 10⁻¹² J/nucleon

4 0
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