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prisoha [69]
3 years ago
12

Pls help asap asap asap aaaaaaa​

Physics
1 answer:
Goryan [66]3 years ago
6 0
Bro the picture is too dark I can’t see do it again so I can help you
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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

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

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PLEASE HELP
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What property of the sound wave determine the volume of sound
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