- #1
Rubidium
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1. Homework Statement
During a cold day, you can warm your hands by rubbing them together. Assume the coefficient of kinetic friction between your hands is 0.500, the normal force between your hands is 35.0 N, and that you rub them together at an average relative speed of 35.0 cm/s.
(a) What is the rate at which mechanical energy is dissipated?
(b) Assume further that the mass of each of your hands is 350 g, the specific heat of your hands is 4.00 kJ/(kg[tex]\cdot[/tex]K), and that all the dissipated mechanical energy goes into increasing the temperature of your hands.
How long must you rub your hands together to produce a 5.00[tex]\circ[/tex]C increase in their temperature?
2. Homework Equations
(a) [tex]\Delta[/tex]E[tex]_{int}[/tex]=Q[tex]_{in}[/tex]+W[tex]_{on}[/tex]
f[tex]_{k}[/tex]=[tex]\mu[/tex][tex]_{k}[/tex]F[tex]_{N}[/tex]
(b) dE[tex]_{int}[/tex]=c[tex]_{v}[/tex]mdT
3. The Attempt at a Solution
(a) f[tex]_{k}[/tex]=[tex]\mu[/tex][tex]_{k}[/tex]F[tex]_{N}[/tex]=(0.500)(35.0 N)=17.5 N [tex]\times[/tex]0.35 m/s=6.125 J/s=6.13 W
(I know that part is correct.)
(b) dE[tex]_{int}[/tex]=c[tex]_{v}[/tex]mdT=(4.00 kJ/(kg[tex]\cdot[/tex]K))(0.35 kg)(5.00[tex]\circ[/tex]C)=7.00 kJ
6.13 J/s / 0.007 J = 876 s
The correct answer is actually 38.1 min = 2286 s but I don't know what I am doing wrong. Please help.
During a cold day, you can warm your hands by rubbing them together. Assume the coefficient of kinetic friction between your hands is 0.500, the normal force between your hands is 35.0 N, and that you rub them together at an average relative speed of 35.0 cm/s.
(a) What is the rate at which mechanical energy is dissipated?
(b) Assume further that the mass of each of your hands is 350 g, the specific heat of your hands is 4.00 kJ/(kg[tex]\cdot[/tex]K), and that all the dissipated mechanical energy goes into increasing the temperature of your hands.
How long must you rub your hands together to produce a 5.00[tex]\circ[/tex]C increase in their temperature?
2. Homework Equations
(a) [tex]\Delta[/tex]E[tex]_{int}[/tex]=Q[tex]_{in}[/tex]+W[tex]_{on}[/tex]
f[tex]_{k}[/tex]=[tex]\mu[/tex][tex]_{k}[/tex]F[tex]_{N}[/tex]
(b) dE[tex]_{int}[/tex]=c[tex]_{v}[/tex]mdT
3. The Attempt at a Solution
(a) f[tex]_{k}[/tex]=[tex]\mu[/tex][tex]_{k}[/tex]F[tex]_{N}[/tex]=(0.500)(35.0 N)=17.5 N [tex]\times[/tex]0.35 m/s=6.125 J/s=6.13 W
(I know that part is correct.)
(b) dE[tex]_{int}[/tex]=c[tex]_{v}[/tex]mdT=(4.00 kJ/(kg[tex]\cdot[/tex]K))(0.35 kg)(5.00[tex]\circ[/tex]C)=7.00 kJ
6.13 J/s / 0.007 J = 876 s
The correct answer is actually 38.1 min = 2286 s but I don't know what I am doing wrong. Please help.