Question

A box sliding on a horizontal frictionless surface encounters a spring attached to a rigid wall and compresses the spring by a certain distance by the time the block ceases its forward motion. If the same experiment is repeated on a rough horizontal surface, some of the box's energy is lost to friction, and so the spring is compressed by a smaller distance by the time the block ceases its forward motion.

Part a) If the mass of the block is 3.96 kg, and its initial velocity is 5 m/s, what is the initial kinetic energy of the block?

________ Joules

Part b) If the block on the frictionless surface compresses the spring by 26.8 cm, what is the spring constant of the spring?

________ N/m

Part c) If the block on the rough surface compresses only compresses the spring by 15 cm, then how much energy was lost to friction?

________ Joules

Part d) If the block traveled a total of 3.20 m on the rough surface between the start of the experiment and the point where the spring is maximally compressed, what is the coefficient of kinetic friction for the block moving on this surface?

Answer 1

Answer:

a) K = 49.5 J  b) k = 1378 N / m c) ΔE = 34 J  d)  μ = 0.399

Explanation:

For this exercise we will use the concepts of energy

a) The initial kinetic energy is

    K = ½ m v²

    K = ½ 3.96 5²

    K = 49.5 J

b) let's use energy conservation

    Em₀ = K = ½ m v²

    $Em_{f}$ = Ke = ½ k x²

    Em₀ =  $Em_{f}$

    ½ m v² = ½ k x²

    k = m v² / x²

    k = 3.96 5² / 0.268²

    k = 1378 N / m

c) Let's calculate the final energy of the spring

     $Em_{f}$ = Ke = ½ k x²

    $Em_{f}$ = ½ 1378 0.15²

     $Em_{f}$ = 15.5 J

The initial energy is the kinetics of the block

    Em₀ = 49.5 J

The lost energy is the difference with the initial

    ΔE =  $Em_{f}$ - Em₀

    ΔE = 15.5 - 49.5

    ΔE = - 34 J

the negative sign means that the energy dissipates

d) For this part we use the concept of work

   W = F d cos θ = ΔK

In this case the force is the friction force that always opposes displacement, so the angle 180 ° and cos 180 = -1

    W = -fr d = ΔK

The force of friction is

   fr = μ N

With Newton's second law

    N-w = 0

    N = W = mg

Let's calculate

   -μ mg d = Kf -K₀o

    μ = K₀ / mgd

    μ = 49.5 / (3.96  9.8  3.20)

    μ = 0.399