Question

The effect of your toss is that the package is moving upward at 10.5 m/s as it leaves your hand. If the 2.5 kg package started at rest and the tossing action lasted 0.39 seconds, what was the net force acting on the package while you were tossing it and what was the strength of the force exerted on the ball by your hand?

Answer 1

@matt101

Answer 2

@Hero

Answer 3

@kropot72

Answer 4

@IrishBoy123

Answer 5

First find acceleration.

delta a = delta v / delta t (delta means the difference essentially)

a = (10.5-0)/0.39
a=25.6m/s^2

F=ma
F=(2.5)(25.6)
F=64N

Answer 6

Okay so is that how you find the net force or the force from the hand?

Answer 7

so the resultant force which was found to be 64N is a summation of the weight and the force by the hand



\(F_h-mg = 64N\)

Answer 8

Okay so this is how you solve for Fnet or the force from the hand?

Answer 9

Because Fh = 39N since you would take (-9.81*2.5)+64

Answer 10

the final force from the weight and the hand combined is equal to 64N because eventually that resultant force is what generates the acceleration.

this 64N is generated by (as i previously told) from the force and the weight (\(F_h-mg\) to be precise)

so \(F_h = mg+64N\) (note that \(F_h\) should be more than 64N )

Answer 11

Okay so the Fh would be 88.5 or 89. So would you take the 89N and subtract it from the 64N to get the net force?

Answer 12

Also sorry about the late reply. I was working on a paper and didn't notice you replied.

Answer 13

net force is the vector summation of the \(F_h\) and \(mg\) which is equal to 64N that causes an acceleration of 25.6ms^(-2)

Answer 14

Okay so I should add the 89N with the 64N to get net force? Is the 89N even right?

Answer 15

@ganeshie8

Answer 16

Okay for the first bit do you take the 10.5 and add it to the 26.92?

Answer 17

The first step to to find the acceleration of the package,

we know the change in velocity, and the change in time,
the acceleration is the ratio of these:
\[a= \frac{\Delta v}{\Delta t}\]

What do you get for this?

Answer 18

Well the Vf would be 10.5m/s and it started from rest so the Vi would be 0. The t is .39s right?

Answer 19

So (10.5-0)/.39

Answer 20

yeah, . .

Answer 21

And get the 26.92 m.s^2

Answer 22

that's right, but the units are m/s^2

Answer 23

Sorry. typo

Answer 24

Now we can calculate the net force on the package, with the mass.
\[F_\text{net}= ma =\]
What do you get for this?

Answer 25

Fnet = 2.5kg * 26.92m/s^2

Answer 26

=67.31N

Answer 27

good,

The net force includes the force of gravity, (which is acting in the opposite direction to the toss)

so the force of the toss must be greater than the net force, by the force of gravity
\[F_\text{toss}=F_\text{net}+F_\text{gravity}\]
where gravitational acceleration \(g= 9.81\,[\text{m/s}^2]\), (assuming you are on earth)

Answer 28

57.50N

Answer 29

the force of the toss should be greater than the net force

Answer 30

So you don't take gravity as a negative?

Answer 31

\[\textbf F_\text{toss}+\textbf F_\text{gravity}=\textbf F_\text{net}\\
F_\text{toss}\hat{\textbf y}+ F_\text{gravity}(-\hat{\textbf y})= F_\text{net}\hat{\textbf y}\\
F_\text{toss}- F_\text{gravity}= F_\text{net}\\
F_\text{toss}= F_\text{net}+ F_\text{gravity}\\
\]

Answer 32

What is the -y?

Answer 33

the gravitational force vector points down,

the toss went up

Answer 34

Okay

Answer 35

So should I get 77.12N then?

Answer 36

what is the magnitude of gravitational force on a 2.5kg object?

Answer 37

No idea

Answer 38

use F = ma, where a = g = 9.81 [m/s^2], m = 2.5 [kg]

Answer 39

Do 24.53

Answer 40

yeah

Answer 41

*So

Answer 42

So \[F_\text{toss}= F_\text{net}+ F_\text{gravity}\\
\qquad= 67.31\,[\text N]+24.52\,[\text N]\\
\qquad=\]

Answer 43

91.84N

Answer 44

Good.

Answer 45

Is the ball the same thing as the package?

Answer 46

No

Answer 47

Well I guess it would be

Answer 48

Okay so the Fnet = 67.31N and the Fhand = 91.84N?

Answer 49

i think so,

Answer 50

So it would be two sig fig since the mass had the least and it was two right?

Answer 51

good point.

Answer 52

Awesome. I got it right. Thank you!!!

Answer 53

I've been curious about this. Is your username from the show?

Answer 54

it isn't

Answer 55

Alright. I've just always wondered that.

Answer 56

Do you have time for another problem by chance? It's fine if you don't.