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1 year ago

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Three objects of the same mass begin their motion at the same height. One object falls straight down, one slides down a low-fric

tion inclined plane, and one swings in a circular arc on the end of a string. All three objects end at the same height.
In which case does the object have the biggest total work done on it by all forces during its motion?
A. Free Fall
B. Incline
C. String
D. Same

1 answer:

erik [133]1 year ago

4 0

Answer:

D. Same

Explanation:

Because only gravity is doing the work on the objects, and gravity is constant for all the objects

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Answer:

y = 54.9 m

Explanation:

For this exercise we can use the relationship between the work of the friction force and mechanical energy.

Let's look for work

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The negative sign is because Lafourcade rubs always opposes the movement

On the inclined part, of Newton's second law

Y Axis

N - W cos θ = 0

The equation for the force of friction is

fr = μ N

fr = μ mg cos θ

We replace at work

W = - μ m g cos θ d

Mechanical energy in the lower part of the embankment

Em₀ = K = ½ m v²

The mechanical energy in the highest part, where it stopped

$Em_{f}$ = U = m g y

W = ΔEm = $Em_{f}$ - Em₀

- μ m g d cos θ = m g y - ½ m v²

Distance d and height (y) are related by trigonometry

sin θ = y / d

y = d sin θ

- μ m g d cos θ = m g d sin θ - ½ m v²

We calculate the distance traveled

d (g syn θ + μ g cos θ) = ½ v²

d = v²/2 g (sintea + myy cos tee)

d = 9.8 12.6 2/2 9.8 (sin16 + 0.128 cos 16)

d = 1555.85 /7.8145

d = 199.1 m

Let's use trigonometry to find the height

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2 years ago

You are testing a new amusement park roller coaster with an empty car with a mass of 130 kg. One part of the track is a vertical

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Answer:

Work done by friction along the motion is given as

$W_f = -5857.8 J$

Explanation:

As per work energy theorem we can say

Work done by all forces = change in kinetic energy of the system

so here car is moving from bottom to top

so here the change in kinetic energy is total work done on the car

so here we will have

$W_f + W_g = \frac{1}{2}m(v_f^2 - v_i^2)$

$W_f - mgH = \frac{1}{2}m(v_f^2 - v_i^2)$

now plug in all data in it

$W_f - (130)(9.81)(2\times 12) = \frac{1}{2}(130)(8^2 - 25^2)$

$W_f = 30607.2 - 36465$

$W_f = -5857.8 J$

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2 years ago

A nonlinear spring is used to launch a toy car. The car is pushed against the spring, compressing the spring 2.5 cm. The force t

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Answer:

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Explanation:

given,

Spring compression, x = 2.5 cm

Force exerts by the spring,

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$PE = \dfrac{1}{2}kx^2$

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Answer:

The speed is $\sqrt{2}v_{0}$ .

(a) is correct option.

Explanation:

Given that,

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We need to calculate the speed

Using formula of initial work done on proton

$W = q V$

We know that,

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$q V=\dfrac{1}{2}mv^2$

Put the value into the formula

$q V_{0}=\dfrac{1}{2}mv_{0}^2$

$v_{0}^2=\dfrac{2qV_{0}}{m}$ ....(I)

If it were accelerated instead through a potential difference of $2 V_{0}$ , then it would gain a speed will be given as :

Using an above formula,

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Put the value of $V_{0}$

$v_{0}'^2=\dfrac{2q\times2V_{0}}{m}$

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$v_{0}'=\sqrt{2}v_{0}$

Hence, The speed is $\sqrt{2}v_{0}$ .

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