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Anvisha [2.4K]
2 years ago
10

01

Physics
1 answer:
klasskru [66]2 years ago
5 0

The statement which describes how a machine can help make work easier is that It can put out more force than the input force by decreasing the distance over which force is applied, therefore the correct option is option A.

<h3>What is work done?</h3>

The total amount of energy transferred when a force is applied to move an object through some distance

The work done is the multiplication of applied force with displacement.

Work Done = Force * Displacement

As work done depends both on the force as well as the displacement force and be reduced by reducing the displacement if the same amount of work is performed by the machine.

The correct answer is option A since the statement that describes how a machine might assist in making work easier says that it can put out greater force than the input force by reducing the distance over which force is exerted.

Learn more about work done from here

brainly.com/question/13662169

#SPJ1

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The diagram illustrating this is shown on the first uploaded

From the question we are told that

   The length of the string is  L =  10.7 \ cm  =  0.107 \ m

     The mass of the bob is  m = 0.344 \  kg

     The angle made  by the string is  \theta  =  5.58^o

The centripetal force acting on the bob is mathematically represented as

         F  =  \frac{mv^2}{r}

Now From the diagram we see that this force is equivalent to

     F  =  Tsin \theta where T is the tension on the rope  and v is the linear velocity  

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          Tsin \theta  =   \frac{mv^2}{r}

Now the downward normal force acting on the bob is  mathematically represented as

          Tcos \theta = mg

So

       \frac{Tsin \ttheta }{Tcos \theta }  =  \frac{\frac{mv^2}{r} }{mg}

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=>   g tan \theta  = \frac{v^2}{r}

The centripetal acceleration which the same as the radial acceleration  of the bob is mathematically represented as

      a_c  =  \frac{v^2}{r}

=>  a_c  = gtan \theta

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     a_c  =  9.8  *  tan (5.58)

     a_c  = 0.9574 m/s^2

The horizontal component is mathematically represented as

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substituting value

   T_x  = 0.344 *  0.9574

    T_x  = 0.3294 \ N

The vertical component of  tension is  

    T_y  =  T \ cos \theta  = mg

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     T_ y  =  0.344 * 9.8

      T_ y  = 3.2712 \ N

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is  

         

       T  = T_x i  + T_y  j

substituting value  

      T  = [(0.3294) i  + (3.3712)j ] \  N

         

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