Question

A student, starting from rest, slides down a water slide. On the way down, a kinetic frictional force (a nonconservative force) acts on her. The student has a mass of 74.0 kg, and the height of the water slide is 11.3 m. If the kinetic frictional force does -5.42 × 103 J of work, how fast is the student going at the bottom of the slide?

Answer 1

To find the student's speed at the bottom of the slide, we apply the conservation of energy principle, accounting for work done by friction. The initial potential energy minus the work done by friction equals the kinetic energy at the bottom. Solving the energy equation for the final velocity will give the required speed.

Step-by-step explanation:

To calculate how fast the student is going at the bottom of the slide, we must apply the principle of conservation of mechanical energy, which states that the total mechanical energy (kinetic plus potential energy) in the absence of nonconservative forces is constant. In this scenario, the mechanical energy is not conserved due to the work done by kinetic friction. We can set up the energy equation as follows:

Potential Energy at the top (PEi) + Kinetic Energy at the top (KEi) + Work done by friction (Wnc) = Potential Energy at the bottom (PEf) + Kinetic Energy at the bottom (KEf)

Since the student starts from rest, KEi = 0, and since the height at the bottom is 0, PEf = 0. Therefore, the equation simplifies to:

mgh + 0 - Wnc = 0 + ½ m v2
74.0 kg × 9.8 m/s2 × 11.3 m - (-5.42 × 103 J) = ½ × 74.0 kg × v2
Solving for v gives us the speed of the student at the bottom of the slide.

Answer 2

The student is going at the bottom of the slide with a velocity of 8.66 m/s

Step-by-step explanation:

Given;

mass of the student, m = 74 kg

height of the water slide, h = 11.3 m

work done, W = -5.42 × 10³ J

Apply work energy theorem;

`W = (1)/(2)mv_f^2+ mgh_f-((1)/(2)mv_o^2 + mgh_o)\\\\ W + (1)/(2)mv_o^2 + mgh_o -mgh_f= (1)/(2)mv_f^2\\\\ W + (1)/(2)mv_o^2 +mg(h_o-h_f) = (1)/(2)mv_f^2\\\\(W)/(m) + (1)/(2) v_o^2 + g(h_o-h_f) = (1)/(2)v_f^2\\\\(2W)/(m)+ v_o^2 + 2g(h_o-h_f) = v_f^2\\\\v_f = \sqrt{(2W)/(m)+ v_o^2 - 2g(h_f-h_o)} \\\\v_f = \sqrt{(2(-5.42*10^3))/(74)+ (0)^2 - 2*9.8(-11.3)}\\\\v_f=√(-146.4865+221.48) \\\\v_f = √(74.9935) \\\\v_f = 8.66 \ m/s`

Therefore, the student is going at the bottom of the slide with a velocity of 8.66 m/s