Question

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 speed of the sled

Answer 1

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

            $134021.99 = 187.5 v^2$

=>         $v = \sqrt{714.783 }$

=>         $v = 26.74 \ m/s$