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Elena-2011 [213]

3 years ago

8

A truck using a rope to tow a 2230-kg car accelerates from rest to 13.0 m/s in a time of 15.0s. How strong must the rope be? μk

= 0.373

1 answer:

Leokris [45]3 years ago

8 0

Answer:

The rope must have a force of 10084,21 N

Explanation

Acceleration calculation

The car acceleration is equal to the acceleration of the truck

ac: car acceleration $\frac{m}{s^{2} }$

at: truck acceleration $\frac{m}{s^{2} }$ )

$ac = at= \frac{vf-vi}{t-ti}$ equation(1)

Known information:

vi = Initial speed = 0, ti = initial time = 0

vf = Final speed = 13 $\frac{m}{s}$ , t = final time =5 s

We replaced the known information in the equation(1):

$ac = at = \frac{13-0}{15-0}$

$ac=ac=\frac{13}{15} \frac{m}{s}$

Dynamic analysis

The forces acting on the car are the following:

Wc: Car weight

N: normal force, road force on the car

Ff: Friction force

T: Force of tension

Car weight calculation:

$Wc=mc*g$

mc = Car mass = 2230kg

g = Gravity acceleration=9.8 $\frac{m}{s^{2} }$

$Wc= 2230*9.8$

$Wc=21854 N$

Normal force calculation:

Newton's first law

$sum Fy= 0$

$N-W=0$

$N=W$

$N=21854 N$

Friction force calculation (Ff):

We have the formula to calculate the friction force:

Ff = μk * N Equation (3)

μk kinetic coefficient of friction

We know that μk = 0.373and N= 21854N ,then:

$Ff=0.373*21854$

$Ff=8151.54 N$

Calculation of the tension force in the rope (T):

Newton's Second law

$sum Fx= mc*ac$

$T-Ff=mc*ac$

$T=2230(\frac{13}{15}) + 8151.54$

T=10084,21 N

Answer: The rope must have a force of 10084,21 N

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1. What are the units of area under the line/curve?<br> 2. Does the area have any meaning?

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The area under the velocity time graph is 125 m and the meaning of the area is displacement.

<h3>What is area under velocity - time graph?</h3>

The area under a velocity time graph represents the displacement of the object.

total area of the graph = A1 + A2

total area of the graph = ¹/₂ (base₁)(height₁) + ¹/₂ (base₂)(height₂)

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A person carries a plank of wood 2.00 m long with one hand pushing down on it at one end with a force f1 and the other hand hold

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Answer:

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⇒ F2 × 0.5 m= 20 kg × 9.81 m/s² × 1 m = 1,962 N×m

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Sergeu [11.5K]

1) Maximum

2) Maximum

Explanation:

The force acting on a mass on a spring is given by Hooke's law; in magnitude:

$F=kx$

where

F is the force

k is the spring constant

x is the displacement

Also we know from Newton's second law that we can write

$F=ma$

where

m is the mass

a is the acceleration

So we can write the equation as

$ma=kx$ (1)

From this relationship, we see that the acceleration is directly proportional to the displacement.

On the other hand, we know that the total mechanical energy of the system mass-spring is constant, and it is given by

$E=\frac{1}{2}kx^2+\frac{1}{2}mv^2=const.$ (2)

where the first term is the elastic potential energy while the second term is the kinetic energy, and where

v is the velocity of the mass

From eq. (2), it is clear that when displacement increases, velocity decreases, and vice-versa; however, from eq.(1) we also know that acceleration is proportional to the displacement.

Therefore this means that:

- When acceleration increases, velocity decreases

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Therefore, the two answers here are:

- When the acceleration of a mass on a spring is zero, the velocity is at a maximum

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