Question

A cord is used to vertically lower an initially stationary block of mass M = 3.6 kg at a constant downward acceleration of g/7. When the block has fallen a distance d = 4.2 m, find (a) the work done by the cord's force on the block, (b) the work done by the gravitational force on the block, (c) the kinetic energy of the block, and (d) the speed of the block. (Note : Take the downward direction positive)

Answer 1

Answer:

a)  W₁ = - 127 J , b) W₂ = 148.18 J, c)   $v_{f}$= 3.43 m/s  and d)  $v_{f}$ = 3.43 m / s

Explanation:

The work is given the equation

         W = F. d

Where the bold indicates vectors, we can also write this expression take the module of each element and the angle between them

        W = F d cos θ

They give us displacement, let's use Newton's second law to find strength, like the block has an equal acceleration (a = g / 7). We take a positive sign down as indicated

       W-T = m a

       T = W -m a

       T = mg -mg/7

       T = mg 6/7

       T = 3.6 9.8 6/7

       T = 30.24 N

Now we can apply the work equation to our problem

a) the force of the cord is directed upwards, the displacement is downwards, so there is a 180º angle between the two

      W₁ = F d cos θ

      W₁ = 30.24 4.2 cos 180

      W₁ = - 127 J

b) the force of gravity is directed downwards and the displacement is directed downwards, the angle between the two is zero (T = 0º)

      W₂ = (mg) d cos 0º

      W₂ = 3.6 9.8 4.2

      W₂ = 148.18 J

c) kinetic energy

      K = ½ m v²

Let's calculate speed with kinematics

    $v_{f}$² = vo² + 2 a y

    v₀ = 0

    a = g / 7

     $v_{f}$² = 2g / 7 y

      $v_{f}$ = √ (2 9.8 4.2 / 7)

      $v_{f}$= 3.43 m/s

We calculate

     K = ½  3.6  3.43²

     K = 21.18 J

d) the speed of the block and we calculate it in the previous part

       $v_{f}$ = 3.43 m / s