Physics, 10.03.2020 08:59 livewithgod

Square root A 1400 kg car is coasting on a horizontal road with a speed of 18 m/s . After passing over an unpaved, sandy stretch 35.0 m long, car's speed has decreased to 14 m/s calculate net force

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Answer from: heyhowdy

The net force acting on the car is equal to 700 Newton.

Given the following data:

Mass of car = 1400 kgInitial speed = 18 m/sFinal speed = 14 m/sDisplacement = 35.0 m

To calculate the net force acting on the car, we would apply the work-energy theorem:

The work-energy theorem

According to the work-energy theorem, the amount of work done by the car must balance the kinetic energy possessed by the car due to its motion.

From the work-energy theorem, we have:

$Work\;done = \Delta K.E\\\\Work\;done = F_{net}\times d=\frac{1}{2} M(V^2-U^2)$

The change in kinetic energy.

$\Delta K.E = \frac{1}{2} M(V^2-U^2)\\\\\Delta K.E = \frac{1}{2} \times 1400 \times (18^2-14^2)\\\\\Delta K.E = 700 \times (324-289)\\\\\Delta K.E = 700 \times 35\\\\\Delta K.E = 24500\;Joules$

Now, we can calculate the net force:

$F_{net} = \frac{\Delta K.E}{displacement} \\\\F_{net} = \frac{24500}{35}$

Net force = 700 Newton.

Read more on work-energy theorem here: link

Answer from: jetcythaop6qlax

The net force on the car is 2560 N.

Explanation:

According to work energy theorem, the amount of work done is equal to the change of kinetic energy by an object. If '' be the work done on an object to change its kinetic energy from an initial value ' $K_{i}$ ' to the final value ' $K_{f}$ ', then mathematically,

$W = K_{f} - K_{i} = \dfrac{1}{2}~m~(v_{f}^{2} - v_{i}^{2})........................................(I)$

where '' is the mass of the object and ' $v_{i}$ ' and ' $v_{f}$ ' be the initial and final velocity of the object respectively. If ' $F_{net}$ ' be the net force applied on the car, as per given problem, and '' is the displacement occurs then we can write,

$W = F_{net}~.~s.......................................................(II)$

Given, $m = 1400~Kg,~v_{i} = 18~m~s^{-1}~v_{f} = 14~m~s^{-1}~and~s = 35~m$ .

Equating equations (I) and (II),

$&& - F_{net} \times 35~m = \dfrac{1}{2} \times 1400~Kg~\times(14^{2} - 18^{2})~m^{2}~s^{-2}\\&or,& F_{net} = \dfrac{\dfrac{1}{2} \times 1400~Kg~\times(14^{2} - 18^{2})}{35}~N\\&or,& F_{net} = 2560~N$

Answer from: Quest

explanation:

as the height from which the ruler was dropped increased, the speed of the ruler at ryan's hand increased.

Answer from: Quest

d₁ = displacement of tern from arctic to antarctic = 11000 miles

d₂ = displacement of tern from antarctic to arctic = - 11000 miles

d = total displacement for the journey

total displacement for the journey is given as

d = d₁ + d₂

d = 11000 + (-11000)

d = 0 miles

t₁ = time of travel from arctic to antarctic = 3 months = 3 x 30 days = 90 days

t₂ = time of travel from antarctic to arctic = 3 months = 3 x 30 days = 90 days

t = total time of travel

total time of travel is given as

t = t₁ + t₂

t = 90 + 90

t = 180 days

average velocity is given as

average velocity = total displacement/ total time

v = d/t

v = 0/180

v = 0 miles/day

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Square root A 1400 kg car is coasting on a horizontal road with a speed of 18 m/s . After passing ov...

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