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2 years ago

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A 2290 kg car traveling at 10.5 m/s collides with a 2780 kg car that is initially at rest at the stoplight. The cars stick toget

her and move 2.80 m before friction causes them to stop. Determine the coefficient of kinetic friction

1 answer:

avanturin [10]2 years ago

7 0

Answer:

0.41

Explanation:

given,

mass of the car, m = 2290 Kg

initial speed = 10.5 m/s

mass of another car, M = 2780 Kg

distance moved = 2.80 m

coefficient of friction = ?

conservation of energy

m u = (M + m) V

2290 x 10.5 = (2290 + 2780) V

V = 4.74 m/s

using equation of motion

v² = u² + 2 a s

4.74² = 2 x a x 2.8

a = 4.02 m/s²

now using equation

a = μ g

4.02 = μ x 9.8

μ = 0.41

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A 54 kg person stands on a uniform 20 kg, 4.1 m long ladder resting against a frictionless wall.

SVETLANKA909090 [29]

A) Force of the wall on the ladder: 186.3 N

B) Normal force of the ground on the ladder: 725.2 N

C) Minimum value of the coefficient of friction: 0.257

D) Minimum absolute value of the coefficient of friction: 0.332

Explanation:

a)

The free-body diagram of the problem is in attachment (please rotate the picture 90 degrees clockwise). We have the following forces:

$W=mg$ : weight of the ladder, with m = 20 kg (mass) and $g=9.8 m/s^2$ (acceleration of gravity)

$W_M=Mg$ : weight of the person, with M = 54 kg (mass)

$N_1$ : normal reaction exerted by the wall on the ladder

$N_2$ : normal reaction exerted by the floor on the ladder

$F_f = \mu N_2$ : force of friction between the floor and the ladder, with $\mu$ (coefficient of friction)

Also we have:

L = 4.1 m (length of the ladder)

d = 3.0 m (distance of the man from point A)

Taking the equilibrium of moments about point A:

$W\frac{L}{2}sin 21^{\circ}+W_M dsin 21^{\circ} = N_1 Lsin 69^{\circ}$

where

$Wsin 21^{\circ}$ is the component of the weight of the ladder perpendicular to the ladder

$W_M sin 21^{\circ}$ is the component of the weight of the man perpendicular to the ladder

$N_1 sin 69^{\circ}$ is the component of the normal force perpendicular to the ladder

And solving for $N_1$ , we find the force exerted by the wall on the ladder:

$N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{mg}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+Mg\frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{(20)(9.8)}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+(54)(9.8)\frac{3.0}{4.1}\frac{sin 21^{\circ}}{sin 69^{\circ}}=186.3 N$

B)

Here we want to find the magnitude of the normal force of the ground on the ladder, therefore the magnitude of $N_2$ .

We can do it by writing the equation of equilibrium of the forces along the vertical direction: in fact, since the ladder is in equilibrium the sum of all the forces acting in the vertical direction must be zero.

Therefore, we have:

$\sum F_y = 0\\N_2 - W - W_M =0$

And substituting and solving for N2, we find:

$N_2 = W+W_M = mg+Mg=(20)(9.8)+(54)(9.8)=725.2 N$

C)

Here we have to find the minimum value of the coefficient of friction so that the ladder does not slip.

The ladder does not slip if there is equilibrium in the horizontal direction also: that means, if the sum of the forces acting in the horizontal direction is zero.

Therefore, we can write:

$\sum F_x = 0\\F_f - N_1 = 0$

And re-writing the equation,

$\mu N_2 -N_1 = 0\\\mu = \frac{N_1}{N_2}=\frac{186.3}{725.2}=0.257$

So, the minimum value of the coefficient of friction is 0.257.

D)

Here we want to find the minimum coefficient of friction so the ladder does not slip for any location of the person on the ladder.

From part C), we saw that the coefficient of friction can be written as

$\mu = \frac{N_1}{N_2}$

This ratio is maximum when N1 is maximum. From part A), we see that the expression for N1 was

$N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}$

We see that this quantity is maximum when d is maximum, so when

d = L

Which corresponds to the case in which the man stands at point B, causing the maximum torque about point A. In this case, the value of N1 is:

$N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{L}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{sin 21^{\circ}}{sin 69^{\circ}}(\frac{W}{2}+W_M)$

And substituting, we get

$N_1=\frac{sin 21^{\circ}}{sin 69^{\circ}}(\frac{(20)(9.8)}{2}+(54)(9.8))=240.8 N$

And therefore, the minimum coefficient of friction in order for the ladder not to slip is

$\mu=\frac{N_1}{N_2}=\frac{240.8}{725.2}=0.332$

Learn more about torques and equilibrium:

brainly.com/question/5352966

#LearnwithBrainly

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3 years ago

a swimmer can swim in still water at a speed of 9.50 m/s. he intends to swim directly across the river that has a downstream cur

Doss [256]

Refer to the diagram shown below.

Still-water speed = 9.5 m/s
River speed = 3.75 m/s down stream.

The velocity of the swimmer relative to the bank is the vector sum of his still-water speed and the speed of the river.

The velocity relative to the bank is
V = √(9.5² + 3.75²) = 10.21 m/s

The downstream angle is
θ = tan⁻¹ 3.75/9.5 = 21.5°

Answer: 10.2 m/s at 21.5° downstream.

7 0

3 years ago

Read 2 more answers

A bullet with a mass 2.25g is fired up into the air with a velocity of 187.5 m/s. What is the maximum height of the bullet

Ksivusya [100]

Answer:

1793.7m

Explanation:

From the principle of conservation of energy; the kinetic energy substended by the object equals the potential energy sustain by the object when it gets to its maximum position.

Now the kinetic energy; is

K.E = 1/2 × m × v2

Where m is mass

v is velocity

Hence.

K.E = 1/2 × 2.25 × (187.5)^2

Now this should be same with the potential energy which is given as;

P.E = m× g× h

Where m is mass of object

g is acceleration of free fall due to gravity = 9.8m/S2

h is maximum height substain by the object.

Hence P.E = 2.25 × 9.8 × h

From the foregoing analysis of energy conversation it implies;

1/2 × 2.25 × (187.5)^2 =2.25 × 9.8 × h

=> 1/2 × (187.5)^2 = 9.8 × h

=>1/2 × (187.5)^2 / 9.8 = h

=> 1793.69m = h

h= 1793.69m

h =1793.7m to 1 decimal place

3 0

3 years ago

a pelican flying along a horizontal path drops a fish from a height of 5.4m. the fish travels 8.0m horizontally before it hits t

oksian1 [2.3K]

Answer:

7.0 m/s

Explanation:

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3 years ago

Calculate the heat, in kilocalories, that is absorbed if 183 g of ice at 0.0 ∘C is placed in an ice bag, melts, and warms to bod

boyakko [2]

Answer:

The total amount of heat needed will be $Q_T=21.411kcal$ .

Explanation:

We will divide the calculation in two: First, the heat needed to melt the ice, and then the heat needed to warm the resulting liquid from 0°C to 37°C.

$m=183g$

$l_f=80\frac{cal}{g} =334\frac{J}{g}$

$l_w=1\frac{cal}{g} =4.184\frac{J}{g}$

<em>i) </em>The fusion heat will be:

$Q_f=l_fm=14640cal=14.640kcal$

<em>ii)</em> The heat needed to warm the water from $T_i=0^{\circ}C$ to $T_i=37^{\circ}C$ will be:

$Q_w=l_wm(T_f-T_i)=6771cal=6.771kcal$

So, the total amount needed will be the sum of these two results:

$Q_T=Q_f+Q_w=14.640kcal+6.771kcal=21.411kcal$ .

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3 years ago