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3 years ago

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What does the slope of the line on this velocity versus time graph represent?

1 answer:

maria [59]3 years ago

8 0

What do you mean when you say "this" graph ? I don't see any graph.

First of all, I'm sure you don't have a "velocity versus time" graph.
You may have a "speed versus time" graph, that doesn't show any
information about the direction of the motion.

As you look at your graph ... a feat that's impossible for me ...

-- If the line is horizontal, then speed is not changing. If the direction
of motion is also not changing, then there is zero acceleration. Sadly,
the graph most likely doesn't carry any information about the direction.
It's possible that the speed is constant but the direction is changing.
Then acceleration isn't zero, but you can't tell that from the graph.

-- If the line is proceeding upward from left to right, whether it's straight
or curving, then there is positive acceleration.

-- If the line is proceeding downward from left to right, whether it's straight
or curving, then there is negative acceleration, or what some people might
call "positive deceleration".

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Physics Question: Determine the work that is being done by tension in pulling the box 193.0 cm along the table. Also determine t

lozanna [386]

1) The work done by tension is 12.0 J

2) The final speed of the box is 1.57 m/s

Explanation:

The work done by a force on an object is given by:

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement

$\theta$ is the angle between the direction of the force and of the displacement

For the box in this problem, we have:

F = 6.2 N is the force applied (the tension in the string)

d = 193.0 cm = 1.93 m is the displacement

$\theta=0$ assuming that the string is horizontal

Substituting, we find the work done by the tension:

$W=(6.2)(1.93)(cos 0)=12.0 J$

2)

In order to determine the final speed of the box, we need to determine its acceleration first.

Beside the tension, acting forward, the other force acting horizontally on the box is the force of friction, whose magnitude is

$F_f = \mu mg$

where

$\mu=0.16$ is the coefficient of friction

m = 2.8 kg is the mass of the box

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

Substituting,

$F_f = (0.16)(2.8)(9.8)=4.4 N$

Therefore, the net force on the box is

$F=6.2 N - 4.4 N = 1.8 N$

And the acceleration can be found by using Newton's second law:

$F=ma$

where

F = 1.8 N is the net force

m = 2.8 kg is the mass of the box

a is the acceleration

Solving for a,

$a=\frac{F}{m}=\frac{1.8}{2.8}=0.64 m/s^2$

Now we can finally find the final speed using the suvat equation:

$v^2-u^2=2as$

where

v is the final speed

u = 0 is the initial speed

$a=0.64 m/s^2$ is the acceleration

s = 1.93 m is the displacement

Solving for v,

$v=\sqrt{u^2+2as}=\sqrt{0+2(0.64)(1.93)}=1.57 m/s$

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