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sergey [27]
3 years ago
5

An object with a mass of 32 kg has an initial energy of 500). At the end of the experimentthe velocity of the object is recorded

as 5.1 m/s . the object travelled 50 m to get to this point, what was the average force of friction on object during the tripAssume no potential energy Show all work
Physics
1 answer:
Alik [6]3 years ago
7 0

Answer:

 F = 1.68 N

Explanation:

Let's solve this exercise in parts.

Let's use the concept of conservation of the mechanical nerve

initial

    Em₀ = 500 J

The energy is totally kinetic

     Em₀ = K = ½ m v₀²

     v₀ = \sqrt{\frac{2 Em_{o} }{m} }

     v₀ = √ (2 500/32)

     v₀ = 5.59 m / s

now with kinematics we can find a space

      v² = v₀² - 2 a x

the negative sign is because the body is stopping

       a =( \frac{v_{o}^{2} - v^{2}   }{2x} )  

let's calculate

       a = (5.59² - 5.1²) / 2 50  

       a = 0.0524 m / s²

Finally let's use Newton's second law

     F = ma

     F = 32 0.0524

     F = 1.68 N

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Answer:

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Explanation:

Hopefully this helped!

4 0
3 years ago
A CFL bulb has an efficiency of 8.9% and a power of 22 W. How much light energy does the lightbulb produce in 1 second
hjlf
The power that the light is able to utilize out of the supply is only 0.089 of the given.
                           Power utilized = (0.089)(22 W)
                                                  = 1.958 W
                                                  = 1.958 J/s
The energy required in this item is the product of the power utilized and the time. That is,
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6 0
4 years ago
Read 2 more answers
Trucks can be run on energy stored in a rotating flywheel, with an electric motor getting the flywheel up to its top speed of 20
Genrish500 [490]

Answer:

The time it can operate between chargins in minutes is

t=102.8 minutes

Explanation:

Given: m=500kg, r=1.0m, w=200\pi rad/s

a). The rotational kinetic energy

K_R=\frac{1}{2}*I*w^2

I=\frac{1}{2}*m*r^2

I=\frac{1}{2}*500kg*(1.0m)^2

I=250 kg*m^2

K_R=\frac{1}{2}*250kg*m^2*(200\pi rad/s)^2

K_R=49.348x10^6J

b). The power average 0.8kW un range time can be find

P=\frac{K_R}{t'}

Solve to t'

t=\frac{K_R}{P}

t=\frac{49.348x10^6}{0.8x10^3w}=6168.5s

t=6168.5s\frac{1minute}{60s}=102.8 minutes

3 0
3 years ago
A uniform plank of length 5.0 m and weight 225 N rests horizontally on two supports, with 1.1 m of the plank hanging over the ri
lawyer [7]

Answer:

x = 0.6034 m

Explanation:

Given

L = 5 m

Wplank = 225 N

Wman = 522 N

d = 1.1 m

x = ?

We have to take sum of torques about the right support point.  If the board is just about to tip, the normal force from the left support will be going to zero.  So the only torques come from the weight of the plank and the weight of the man.

∑τ = 0  ⇒     τ₁ + τ₂ = 0  

Torque come from the weight of the plank = τ₁

Torque come from the weight of the man = τ₂

⇒  τ₁ = + (5 - 1.1)*(225/5)*((5 - 1.1)/2) - (1.1)*(225/5)*((1.1)/2) = 315 N-m (counterclockwise)

⇒  τ₂ = Wman*x = 522 N*x   (clockwise)

then

τ₁ + τ₂ = (315 N-m) + (- 522 N*x) = 0

⇒  x = 0.6034 m

7 0
3 years ago
A wheel of mass 4kg is pulled up a plane inclined at 30° to the horizontal by a force of 45N applied to the axle and parallel to
Len [333]

Answer:

v = 10 m/s

Explanation:

Let's assume the wheel does not slip as it accelerates.

Energy theory is more straightforward than kinematics in my opinion.

Work done on the wheel

W = Fd = 45(12) = 540 J

Some is converted to potential energy

PE = mgh = 4(9.8)12sin30 = 235.2 J

As there is no friction mentioned, the remainder is kinetic energy

KE = 540 - 235.2 = 304.8 J

KE = ½mv² + ½Iω²

ω = v/R

KE = ½mv² + ½I(v/R)² = ½(m + I/R²)v²

v = √(2KE / (m + I/R²))

v = √(2(304.8) / (4 + 0.5/0.5²)) = √101.6

v = 10.07968...

5 0
3 years ago
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