A child whose weight is 267 N slides down a 6.10 m long playground slide that makes an angle of 20° with the horizontal. The coefficient of kinetic friction between slide and child is 0.10. (a) How much energy is transferred to thermal energy? (b) If she starts at the top with a speed of 0.457 m/s, what is her speed at the bottom?

(a) ΔE = 153.05 J

(b) Her speed at the bottom is 5.465 m/s

Explanation:

Here, we have

Slope of incline = 20 °

Normal force to incline = mg cos θ = F cos θ = 267 × cos 20 = 250.9 N

Force of friction = Kinetic friction × Normal force = 0.1 × 250.9 N  = 25.09 N

If all the work done by the child sliding down the inclined plane is converted into thermal energy, we have

Work done = 25.09 N × 6.10 m = 153.05 J

The heat energy transferred to thermal energy. ΔE  = 153.05 J

(b) Here we have the some of the energies at the bottom is given by

ΔU + ΔK.E + ΔE = 0

Where:

ΔU = Gravitational potential energy

ΔK.E = Change in kinetic energy of the child

ΔE  = Change in thermal energy of the child

ΔU = m·g·h

Where:

m·g = Weight of the child = 267 N and

h = Height from which the child starts the slide = Slope height

Slope height = 6.1 × sin 20° = 2.09 m

∴ ΔU = m·g·h = 267 N × 2.09 m = 557.05 J

Therefore. from ΔU + ΔK.E + ΔE = 0, we have

-557.05 J + ΔK.E + 153.05 J = 0

ΔK.E = 557.05 J - 153.05 J = 404.0 J

However, ΔK.E = (0.5×m×(v₂²-v₁²)

Where,

v₁ = Initial velocity of child = 0.457 m/s

v₂ = Final velocity of child

Where 267 = m×g

Where:

g = Acceleration due to gravity = 9.8 m/s²

m = Mass of child

m = 267/g = 267/9.8 = 27.245 kg

404.0 J = (0.5×27.245×(v₂²-0.457²)

29.66 = v₂²-0.457²

v₂² = 29.66 + 0.457² = 29.866 m²/s²

v₂  = 5.465 m/s

Her speed at the bottom = 5.465 m/s.

the area between the graph and the t axis