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Oxana [17]
3 years ago
14

If the sled is also being acted on by a constant kinetic friction force of 260 N and it started from rest, what is the final spe

ed of the sled
Physics
1 answer:
larisa86 [58]3 years ago
6 0

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

  W = 121271.99 \  J

b

  v  = 26.74 \  m/s

Explanation:

From the question we are said that

        The  mass of the shed is  m_s  =  375 \  kg

        The pulling  force is  F =  2511 \  N

        The  angle is  \theta  =  15^o

        The  distance covered is  d =  50 \ m

        The kinetic force is  F_k  =  255 \  N

Generally the work done is mathematically represented as

           W =  F * cos(\theta ) * d

=>         W = 2511 * cos( 15  ) *  50

=>          W = 121271.99 \  J

Generally according to the work-energy  theorem

           T_w =  \Delta  KE

Here  T_w is total work done which is mathematically represented as

       T_w =  F_k *  d  +  F d cos(\theta)

=>    T_w =   255  *   50   + 121271.99

=>    T_w =   134021.99 \  J

  Also  \Delta  KE  is the change in kinetic energy which is mathematically represented as

            \Delta  KE  =  \frac{1}{2}  *  m *  [ v^2 -  u^0 ]

Here u  is the initial velocity of the shed when at rest

           \Delta  KE  =  \frac{1}{2}  * 375  *  [ v^2 -  0^0 ]

=>      \Delta  KE  =  187.5  v^2

So

            134021.99  =   187.5  v^2

=>         v  =  \sqrt{714.783 }

=>         v  = 26.74 \  m/s

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