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Romashka-Z-Leto [24]

3 years ago

7

A small piece of Styrofoam packing material is dropped from a height of 2.60 m above the ground. Until it reaches terminal speed

, the magnitude of its acceleration is given by a = g − Bv. After falling 0.600 m, the Styrofoam effectively reaches terminal speed, and then takes 4.50 s more to reach the ground.
(a) What is the value of the constant B? s-1.
(b) What is the acceleration at t = 0? [m/s2 (down)].
(c) What is the acceleration when the speed is 0.150 m/s? [m/s2 (down)].

1 answer:

Vadim26 [7]3 years ago

7 0

Answer:

Explanation:

a ) After the attainment of terminal speed , object takes 4.5 s to cover a distance of 2 m

So terminal speed V = 2 / 4.5

= .444 m /s

When it attains terminal speed , acceleration becomes zero

0 = g - B x .444

B = 22.25 s⁻¹

b ) At t = 0 , v = 0

a = g - B v

a = g at t = 0

c ) When v = .15

a = g - 22.25 x .15

= 9.8 - 3.31

= 6.5 m /s²

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Answer:

The rope must have a force of 10084,21 N

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Acceleration calculation

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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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