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3 years ago

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A backpack has a mass of 8 kg. It is lifted and given 54.9 J of gravitational potential energy. How high is it lifted? Accelerat

ion due to gravity is g = 9.8 m/s2.

2 answers:

sweet [91]3 years ago

8 0

Potencial Energy=hma
Where
Potencial Energy =E= 54.9J
h=?
m=8kg
a=9.8m/s^2
You need to know that 1 J=1(kgm^2)/s^2

Isolate h=E/(ma)
h=(54.9)/(8*9.8)

soldier1979 [14.2K]3 years ago

4 0

Answer: The potential energy for a lifted object is U = m*a*h

where m is the mass, h is the height and a is the acceleration.

in our problem the mass is m = 8kg, the height is the thing we want to find, a is the gravitational acceleration, 9.8 m/s^2 and U = 54.9j

then our problem is:

54.9j = h*9.8*8 kg*m/s^2

then h = 54.9/(9.8*8) = 0.7m

Then the backpack is lifted 0.7 meters.

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Answer: 37.981 m/s

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the ball has two components: <u>x-component</u> and <u>y-component.</u> Being their main equations as follows:

<u>x-component: </u>

$x=V_{o}cos\theta t$ (1)

Where:

$x=52 m$ is the point where the ball strikes ground horizontally

$V_{o}$ is the ball's initial speed

$\theta=0$ because we are told the ball is thrown horizontally

$t$ is the time since the ball is thrown until it hits the ground

<u>y-component: </u>

$y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=120m$ is the initial height of the ball

$y=0$ is the final height of the ball (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity

Knowing this, let's start by finding $t$ from (2):

<u></u>

$0=y_{o}+V_{o}sin(0\°) t+\frac{gt^{2}}{2}$ (3)

$0=y_{o}+\frac{gt^{2}}{2}$

$t=\sqrt{\frac{-2 y_{o}}{g}}$ (4)

$t=\sqrt{\frac{-2 (120 m)}{-9.8m/s^{2}}}$ (5)

$t=4.948 s$ (6)

Then, we have to substitute (6) in (1):

$x=V_{o}cos(0\°) t$ (7)

And find $V_{o}$ :

$V_{o}=\frac{x}{t}$ (8)

$V_{o}=\frac{52 m}{4.948 s}$ (9)

$V_{o}=10.509 m/s$ (10)

On the other hand, since we are dealing with constant acceleration (due gravity) we can use the following equation to find the value of the ball's final velocity $V$ :

$V=V_{o} + gt$ (11)

$V=10.509 m/s + (-9.8 m/s^{2})(4.948 s)$ (12)

$V=-37.981 m/s$ (13) This is the ball's final velocity, and the negative sign indicates its direction is downwards.

However, we were asked to find the <u>ball's final speed</u>, which is the module of the ball's final vleocity vector. This module is always positive, hence the speed of the ball just before it strikes the ground is 37.981 m/s (positive).

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