Question

A child whose weight is 287 N slides down a 7.20 m playground slide that makes an angle of 31.0° with the horizontal. The coefficient of kinetic friction between slide and child is 0.120. (a) How much energy is transferred to thermal energy? (b) If she starts at the top with a speed of 0.559 m/s, what is her speed at the bottom?

Answer 1

Answer:

a

$H =212.6 \ J$

b

$v = 7.647 \ m/s$

Explanation:

From the question we are told that

   The child's weight is  $W_c = 287 \ N$

    The length of the sliding surface of the playground is  $L = 7.20 \ m$

    The coefficient of friction is  $\mu = 0.120$

      The angle is $\theta = 31.0 ^o$

      The initial  speed is  $u = 0.559 \ m/s$

Generally the normal force acting on the child is mathematically represented as

=>    $N = mg * cos \theta$

Note  $m * g = W_c$

Generally the frictional force between the slide and the child is    

         $F_f = \mu * mg * cos \theta$

Generally the resultant force acting on the child due to her weight and the frictional  force is mathematically represented as

      $F =m* g sin(\theta) - F_f$

Here  F is the resultant force and it is represented as  $F = ma$

=>   $ma = m* g sin(31.0) - \mu * mg * cos (31.0)$

=>   $a = g sin(31.0)- \mu * g * cos (31.0)$

=>  $a = 9.8 * sin(31.0) - 0.120 * 9.8 * cos (31.0)$

=>$a = 4.039 \ m/s^2$

So

   $F_f = 0.120 * 287 * cos (31.0)$

=> $F_f = 29.52 \ N$

Generally the heat energy generated by the frictional  force which equivalent tot the workdone by the frictional force  is mathematically represented as

     $H = F_f * L$

=>  $H = 29.52 * 7.2$

=>  $H =212.6 \ J$

Generally from kinematic equation we have that

    $v^2 = u^2 + 2as$

=>  $v^2 = 0.559^2 + 2 * 4.039 * 7.2$

=>  $v = \sqrt{0.559^2 + 2 * 4.039 * 7.2}$

=>  $v = 7.647 \ m/s$