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4 years ago

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Physics

1 answer:

lys-0071 [83]4 years ago

4 0

Answer:

The object will have an upward acceleration

Explanation:

Let's consider the forces applied on the box. We have only two forces:

- The applied force of push, $F_p$ , downward

- The force of gravity, $W$ , (also known as weight of the object), downward

Therefore, the net force on the box is:

$F_{net}=F_p -W$

Here, we know that force applied is equal or greater than the weight, so

$F_p \geq W$

And therefore the net force is greater than zero:

$F_{net}\geq 0$ (1)

According to Newton's second law of motion, the net force on the box is equal to the product between its mass and its acceleration:

$F_{net}=ma$

where

m is the mass of the box

a is its acceleration

Given (1), this means that

$a\geq 0$

Therefore, the box will have an upward acceleration.

In this case force example we have:

$F_p = 100 N\\W = 40 N$

So the mass of the box is

$m=\frac{W}{g}=\frac{40}{10}=4 kg$

So the net force is

$F_{net}=F_p-W=100-40=60 N$

And the acceleration is

$a=\frac{F_{net}}{m}=\frac{60}{4}=15 m/s^2$

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Suppose the ball is thrown from the same height as in the PRACTICE IT problem at an angle of 27.0°below the horizontal. If it st

andreev551 [17]

a) The time of flight is 3.78 s

b) The initial speed is 17.0 m/s

c) The speed at impact is 46.4 m/s at $70.3^{\circ}$ below the horizontal

Explanation:

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The motion of the ball is a projectile motion, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction

We start by considering the vertical motion, to find the time of flight of the ball. We do it by using the following suvat equation: for the y-displacement:

$y=u_y t+\frac{1}{2}at^2$

where we have:

y = -45.0 m is the vertical displacement of the ball (the height of the building)

$u_y=u sin \theta$ is the initial vertical velocity, with u being the initial velocity (unknown) and $\theta=-27.0^{\circ}$ the angle of projection

t is the time of the fall

$a=g=-9.8 m/s^2$ is the acceleration of gravity

Along the x-direction, the equation of motion is instead

$x=(u cos \theta)t$

where $ucos \theta$ is the horizontal component of the velocity. Rewriting this equation as

$t=\frac{x}{ucos \theta}$

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$y=xtan \theta + \frac{1}{2}gt^2$

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And solving for t, we find the time of flight:

$t=\sqrt{\frac{y-x tan \theta}{g}}=\sqrt{\frac{-45-(59.0)(tan(-27^{\circ}))}{-9.8}}=3.78 s$

We can now find the initial speed, u, by using the equation of motion along the x-direction

$x=u cos \theta t$

where we know that:

x = 59.0 m is the horizontal range

$\theta=-23^{\circ}$ is the angle of projection

$t=3.78 s$ is the time of flight

Solving for u, we find the initial speed:

$u=\frac{x}{cos \theta t}=\frac{59.0}{(cos (-23^{\circ}))(3.78)}=17.0 m/s$

First of all, we notice that the horizontal component of the velocity remains constant during the motion, and it is equal to

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The vertical velocity instead changes according to the equation

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Substituting all the values and t = 3.78 s, the time of flight, we find the vertical velocity at the time of impact:

$v_y = (17.0)(sin (-23^{\circ}))+(-9.8)(3.78)=-43.7 m/s$

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