Answered

Explore IDNLearn.com's extensive Q&A database and find the answers you need. Join our knowledgeable community and get detailed, reliable answers to all your questions.

A 200g block on a 50cm long string swings in a circle, it's frictionless and 75rpm. What is its speed and tension on string

Sagot :

Jackattackbluntidn
angular velocity = (75x2pie)/60
                          =2.5pie ras^-1 
linear velocity(or speed) at end of string, v = radius x angular velocity
                                                           v= 0.5 x 2.5pie
                                                           v=3.93 ms^-1

tension of string (I beleve is centeral force aplied by string), F= (mv^2)/r
                                                                                      F= (0.2 x 3.93^2)/0.5
                                                                                      F=6.18 N
(sorry if wrong)
Carlosegoidn

The speed of the block is given by:

[tex] V = w * R
[/tex]

Where,

w: angular speed

r: radius of the circular path.

The angular velocity must be in radians over seconds:

[tex] w = (75) * (2\pi) * (\frac{1}{60})

w = 7.85
[/tex]

The radius must be in the subway:

[tex] R = (50) * (\frac{1}{100})

R = 0.5 m
[/tex]

Then, the speed is given by:

[tex] V = (7.85) * (0.5)

[/tex]

[tex] V = 3.925 \frac{m}{s}
[/tex]

The tension of the rope is the centripetal force.

By definition, the centripetal force is:

[tex] F = m * (\frac{V^2}{R})
[/tex]

Where,

m: mass of the block in kilograms

Substituting values:

[tex] F = 0.2 * (\frac{3.925 ^ 2}{0.5})

F = 6.2 N
[/tex]

Answer:

its speed and tension on string are:

[tex] V = 3.925 \frac{m}{s}

F = 6.2 N [/tex]

We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Your questions deserve reliable answers. Thanks for visiting IDNLearn.com, and see you again soon for more helpful information.