Question

A child whose weight is 235 N slides down a 4.90 m playground slide that makes an angle of 37.0° with the horizontal. The coefficient of kinetic friction between slide and child is 0.0510. (a) How much energy is transferred to thermal energy? (b) If she starts at the top with a speed of 0.355 m/s, what is her speed at the bottom?

Answer 1

Answer:

Explanation:

a) First, let's calculate the value of the Friction force, which is given by the formula:

Ff = u*W

As the Friction force has an X component, it would be:

Ff = u*m*g*cosФ

Where m*g is the weight of the child.

Solving for Ff:

Ff = 0.051 * 235cos37 = 9.57 N

Now, to get the energy transferred to thermal energy (or heat) we need to get the Work done, so:

Wf = Ff * d

Wf = 9.57 * 4.9 = 46.9 J

b) We need to get the downslope component of the child weight, which is:

Wy = 235sin37 = 141.43 N

As you can imagine, the gravity also does work, so:

Wg = Wy * d

Wg = 141.43 * 4.9 = 693 J

Now, let's get the kinetic energy, which can be obtained by this expression:

ΔKe = Wg - Wf

ΔKe = 693 - 46.9 = 646.1 J

The formula for kinetic energy is:

Ke = 1/2 m*V^2

We have the innitial speed which is 0.355 m/s, and the mass can be obtained by m*g so:

Fw= m*g ----> m = Fw/g

m = 235 / 9.8 = 23.97 kg

so the innitial energy is:

Ke = 1/2 * 23.97 * (0.355)^2

Ke = 1.51 J

This could be Ke1, so to get Ke2:

ΔKe = Ke2 - Ke1

Ke2 = Δke + Ke1

Ke2 = 646.1 + 1.51 = 647,61 J

Finally, the speed at the bottom would be:

v = √2Ke/m

v = √2*647.61/23.97

v = 7.35 m/s