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.

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz