Question

A 3.0-kg block is dragged over a rough horizontal surface by a constant force of 16 N acting at an angle of 37° above the horizontal as shown. The speed of the block increases from 2 m/s to 3.8 m/s in a displacement of 5.5 m. What work was done by the friction force during this displacement?

Answer 1

Answer:

The  value  is   $W = -54.615 \ J$

Explanation:

From the question we are told that

    The mass of the block is  $m = 3.0 \ kg$

     The force is $F = 16 \ N$

     The  angle is  $\theta = 37^o$

     The first speed of the block is  $u = 2 \ m/s$

     The second  speed of the block is  $v = 3.8 \ m/s$

     The displacement is  $d = 5.5 \ m$

Gnerally from kinematic equation we have that

      $v^2 = u^2 + 2as$

=>    $3.8 ^2 = 2^2 + 2 * a* 5.5$

=>    $a = 0.9491 \ m/s^2$

Generally the net force acting on the crate is mathematically represented as

        $F_{net} = [ F cos (\theta ) - F_f ] = ma$

Here  $F_f$ is the frictional force acting on the crate

So    

          $[ 16 cos (37 ) - F_f ] = 3 * 0.9491$    

=>       $F_f = 9.93 \ N$

Generally the work done by  friction  during the displacement is mathematically represented as

        $W = F_f * d * cos ( 180 )$

=>     $W = 9.93 * 5.5 * cos ( 180 )$

=>     $W = -54.615 \ J$