Physics Problems & Solutions: #06

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Classical Mechanics - Friction Force

A block of mass m sliding down an incline at constant speed is initially at height h above the ground as shown in the figure. The coefficient of kinetic friction between the mass and the incline is µ. If the mass continues to side down the incline at a constant speed, how much energy is dissipated by friction by the time the mass reached the bottom of the incline?

A. mgh/µ
B. mgh
C. µmgh/sin θ
D. mgh sin θ
E. 0

(GR9677 #06)

Solution:

mg sin θ − F_r= ma
At a constant speed, a = 0
F_r= mg sin θ

Energy dissipated = Work done by the frictional force
W = F_r⋅ s

s = length of the inclined surface = h/sin θ

W = mg sin θ (h/sin θ) = mgh

Answer: B

Classical Mechanics - Friction Force

Two wedges, each of mass m, are placed next to each other on a flat floor. A cube of mass M is balanced on the wedges as shown above. Assume no friction between the cube and the wedges, but a coefficient of static friction $\mu<1$ between the wedges and the floor. What is the largest M that can be balanced as shown without motion of the wedges?

A. $\frac{m}{\sqrt2}$

B. $\frac{\mu m}{\sqrt2}$

C. $\frac{\mu m}{1-\mu}$

D. $\frac{2\mu m}{1-\mu}$

E. All M will balance

(GR9277 #06)

Solution:

$2N_w=2mg+Mg\rightarrow N_w=mg+\frac{Mg}{2}$

$\begin{aligned} \\F_x&=f \\N_c \cos 45&=\mu N_w \\Mg (\cos 45)^2&=\mu \left (mg+\frac{Mg}{2} \right ) \\M \left (\frac{\sqrt{2}}{2} \right )^2&=\mu m+\frac{\mu M}{2} \\\frac{M}{2} &=\mu m+\frac{\mu M}{2} \\M &=2\mu m+2\mu M \\M &=\frac{2\mu m}{1-\mu} \end{aligned}$

Answer: D

Classical Mechanics - Kinematics

A particle is initially at rest at the top of a curved frictionless track. The x- and y-coordinate of the track are related in dimensionless units by y = x²/4, where the positive y-axis is in the vertical downward direction. As the particle slides down the track, what its tangential acceleration?

A. 0
B. g
C. gx/2
D. gx/√(x²+4)
E. (gx²)/√(x²+16)

(GR8677 #06)

Solution:

A. FALSE.
The particle slides down the curved track → the tangential acceleration is not zero.

B. FALSE.
The particle slides down the curved track, not in the vertical downward direction (the direction of gravity acceleration) → the tangential acceleration is not equal to g.

C. FALSE.
The unit of gx/2 is not the unit of acceleration.

D. TRUE.
The unit of gx/√(x²+4) = g → the unit of acceleration.

E. FALSE.
The unit of (gx²)/√(x²+16) is not the unit of acceleration.

Answer: D

Calculation:

$\\a_\text{tan}=g \sin \alpha\\\\\tan \alpha = \frac{dy}{dx}=\frac{x}{2}$

$\\\frac{\sin \alpha}{\cos \alpha}=\frac{\sin \alpha}{\sqrt{1-\sin^2\alpha}}=\frac{x}{2}\\\\\rightarrow \frac{\sin^2 \alpha}{1-\sin^2\alpha}=\frac{x^2}{4}\\\\\rightarrow4\sin^2 \alpha=x^2-x^2\sin^2 \alpha\\\rightarrow\sin^2 \alpha=\frac{x^2}{4+x^2}\\\rightarrow \sin \alpha=\frac{x}{\sqrt{4+x^2}}$

$\rightarrow a_{\tan}=\frac{gx}{\sqrt{4+x^2}}$

Thermal Physics - Isothermal vs Adiabatic

An ideal monatomic gas expands quasi-statically to twice its volume. If the process is isothermal, the work done by the gas is W_i. If the process is adiabatic, the work done by the gas is W_a. Which is the following is true?

A. W_i= W_a
B. 0 = W_i

W_a
C. 0

W_i

W_a
D. 0 = W_a

W_i
E. 0

W_a

W_i

(GR0177 #06)

Solution:

Isothermal and Adiabatic P-V Diagram

Adiabatic connects high-T isotherm and low-T isotherm.
Isothermal line is always higher than the adiabatic line and they both end at the same volume
The area under the isothermal line is bigger than the adiabatic → W_a W_i

Answer: E

Calculation:

Isothermal:

PV = constant
P_iV_i= P_fV_f= constant
Given V_f= 2V_i

P_iV_i= 2P_fV_i

P_f (iso)= ½ P_i

Adiabatic:

PV^γ= c
P_iV_i^γ= P_fV_f^γ= constant
V_f= 2V_i
P_iV_i^γ= P_f(2V_i)^γ= 2^γP_fV_i^γ
P_f (adi)= (1/2^γ)P_i

Therefore,

P_{f (iso)}/P_{f (adi)}= 2^γ/ 2
P_f (iso)= 2^γ⁻¹P_{f (adi)}
→ P_adi

P_iso
W = ∫ PdV → W_adi

W_iso