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3 years ago

5

While on a playground, you and your niece take turns sliding down a frictionless slide. Your mass is 75 kg while your little nie

ce's mass is only 25 kg. Assume that both of yo begin sliding from rest from the same height. Which one of the following statements b describes who has a larger speed at the bottom of the slide?
a. You, because your greater weight causes a greater downward acceleration.
b. Your niece, because she is not pressing down against the side as strongly her motion is closer to free fall than yours.
c. Both of you have the same speed at the bottom.
d. Your niece, because lighter objects are easier to accelerate.
e. You, because you take less time to slide down.

1 answer:

Olenka [21]3 years ago

6 0

Answer:

c.

Explanation:

In absence of friction, total mechanical energy must be conserved.
This means that the change in gravitational potential energy (in magnitude) must be equal to the change in kinetic energy:

$\Delta K = \Delta U \\ \\ \frac{1}{2} * m* v^{2} = m*g*h$

As it can be seen, the mass m is on the both sides of the equation, which means that it can be simplified.
We can solve this equation for v (speed at the bottom) as follows:

$v = \sqrt{2*g*h}$

As it can be seen, the mass has no part in the equation, so, due both are starting from the same height, both people have the same speed at the bottom.
The option c is the right one.

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Read 2 more answers

A ball is thrown with an initial speed vi at an angle i with the horizontal. The horizontal range of the ball is R, and the ball

adell [148]

Answer:

Part a)

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

$v_y = \sqrt{Rg/3}$

Part d)

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

$\theta_i = 33.7 degree$

Part f)

$H = \frac{13R}{8}$

Part g)

$X = \frac{13R}{4}$

Explanation:

Initial speed of the launch is given as

initial speed = $v_i$

angle = $\theta_i$ degree

Now the two components of the velocity

$v_x = v_i cos\theta_i$

similarly we have

$v_y = v_i sin\theta_i$

Part a)

Now we know that horizontal range is given as

$R = \frac{v_i^2 (2sin\theta_icos\theta_i)}{g}$

maximum height is given as

$H = \frac{R}{6} = \frac{v_i^2 sin^2\theta_i}{2g}$

so we have

$v_i sin\theta = \sqrt{Rg/3}$

time of flight is given as

$T = \frac{2v_isin\theta_i}{g}$

$T = \frac{2\sqrt{Rg/3}}{g}$

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

Now the speed of the ball in x direction is always constant

so at the peak of its path the speed of the ball is given as

$R = v_x T$

$R = v_x 2\sqrt{\frac{R}{3g}}$

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

Initial vertical velocity is given as

$v_y = v_i sin\theta_i$

$v_i sin\theta = \sqrt{Rg/3}$

Part d)

Initial speed is given as

$v = \sqrt{v_x^2 + v_y^2}$

so we will have

$v = \sqrt{Rg/3 + 3Rg/4}$

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

Angle of projection is given as

$tan\theta_i = \frac{v_y}{v_x}$

$tan\theta_i = \frac{\sqrt{Rg/3}}{\sqrt{3Rg}/2}$

$\theta_i = 33.7 degree$

Part f)

If we throw at same speed so that it reach maximum height

then the height will be given as

$H = \frac{v^2}{2g}$

$H = \frac{13R}{8}$

Part g)

For maximum range the angle should be 45 degree

so maximum range is

$X = \frac{v^2}{g}$

$X = \frac{13R}{4}$

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3 years ago