Physics 101 - Homework # 4 Solutions

• Problem 5-6
• Problem 5-10
• Problem 5-14
• Problem 5-20
• Problem 5-28
• Problem 5-32
• Problem 5-36
• Problem 5-46
• Problem 5-49
• Problem 6-5
• Problem 6-6
• Problem 6-45
• Problem 6-83

Note: Symbols written in Bold are vectors.

Chapter 5

Problem 6)

The tension in the rope is created by the contact between the rope and the man's hands. The man will pull down on the rope, and the rope will pull up on the man. As shown in the figure, we take down to be the positive direction (this is the direction of the acceleration).

a) For the man we can write

F_y = ma_y
mg - T = ma

From this we see that there is a minimum acceleration that produces the maximum tension:

(85 kg)(9.8 m/s²) - 700 N = (85 kg)a_min

which gives

a_min = 1.56 m/s²

b) For this constant acceleration, we can write

y = y₀ + v₀t + ½at²

= 0 + 0 + ½(1.6 m/s²)(4.0 s)²

= 12.5 m

To find the velocity, we use

v = v₀ + at = 0 + (1.6 m/s²)(4.0 s) = 6.24 m/s   down.

Problem 10)

a) If we choose the two blocks as a system we can write F = ma for the x component

F = (m₁ + m₂)a which gives

a = F/(m₁ + m₂)

b) Choosing the block on the right as the system now, then F = ma gives, for the x-component,

F₂= m₂a = m₂F/(m₁ + m₂)

Note that an equal and opposite force acts on m₁.

Problem 14)

There are two different coodinate systems used, one for each of the blocks. Since the string doesn't stretch, the tension is the same at each end of the string.

Now write F = ma for each block, using the force diagrams:

Block 1, x-component: m₁g sin₁ - T = 0

Block 2, x-component: T - m₂g sin₂ = 0

since we require the acceleration to be 0. Note that we don't need the y-component equations here. Adding the two equations together, to eliminate T, we get

m₁g sin₁ = m₂g sin₂

Plugging in numbers we get

(1.50 kg)sin62° = (2.50 kg)sin₂

which gives

₂ = 32°

Problem 20)

The acceleration can be found from the boy's one-dimensional motion:

v = v₀ + at

0 = 4.5 m/s + a(3.4 s)

which gives

a = - 1.32 m/s²

Using the force diagram for the boy, we can write F = m a

x-component: - µ_kF_N = Ma

y-component: F_N - Mg = 0

Thus

F_N = Mg

and

µ_k = - a/g

= - (-1.32 m/s²)/(9.8 m/s²) = 0.135

Problem 28)

Static friction must be backward, to oppose the impending slide along the truck bed and provide the deceleration of the crate. We can find the acceleration from the motion of the truck:

v² = v₀² + 2a(x-x₀)

0 = (26.7 m/s)² + 2a(140 m -0)

which gives

a = - 2.55 m/s²

We write   F = ma   from the force diagram for the crate:

x-component: -f_s = ma

y-component: F_N - Mg = 0

Thus

f_s = -ma = -(250 kg)(-2.6 m/s²) = 638 N

and

F_N = Mg = (250 kg)(9.8 m/s²) = 2450 N

Because   f_s = f_s,max = µ_sF_N   the minimum µ required is

µ_s,min = f_s/F_N = (638 N)/(2450 N) = 0.26

Problem 32)

The two blocks must have the same acceleration; we apply F = ma for the force diagram for the upper block:

x-component: F - f = m₁a

y-component: F_N1 - m₁g = 0

Writing F = ma for the lower block:

x-component: f = m₂a

y-component: F_N2 - F_N1 - m₂g = 0

From the y-component equations we have

F_N1 = m₁g    and   F_N2 = (m₁+m₂)g

We can eliminate a from the two x-equations, to get

F = (m₁ + m₂)f/m₂

The maximum value of F is therefore when f is at its maximum value, which is

f_max = µ_sF_N1 = µ_sm₁g

Thus

F_max = (1.2 kg + 1.8 kg)[0.3(1.g kg)(9.8 m/s²]/(1.8 kg)

= 5.9 N

If the force F is applied to the lower block, the x-equations now become

top block: f = m₁a

bottom block: F - f = m₂a

We can eliminate a from these equations, to yield

F = (m₁ + m₂)f/m₁

so the maximum value of F becomes

F_max = (1.2 kg + 1.8 kg)[0.3(1.g kg)(9.8 m/s²]/(1.2 kg)

= 8.8 N

Problem 36)

At terminal velocity, a = 0 (by definition!), and the weight is balanced by the drag force:

F_drag = mg = kv²

so

k = mg/v² = 116 kg(9.8 m/s²)/(4.9 m/s²) = 4.7 Ns²/m² = 4.7 kg/m

Problem 46)

The centripetal acceleration is

a_r = v²/r = (200 m/s)²/(30 x 10³ m) = 1.3 m/s²     toward the center

Problem 49)

If the automobile does not skid, the friction is static, with f_s <= µ_sF_N. At high speed f_s will be down the incline. Note that we take a coordinate system with the x-axis in the direction of the centripetal acceleration.

We write F = ma from the force diagram for the car :

x-component: f_Nsin + f_scos = ma = mv²/R

y-component: F_Ncos - f_ssin - mg = 0

The speed is maximum when   f_s = f_s,max = µ_sN

From the y-equation we get:

F_Ncos - µ_sF_Nsin = mg

or

F_N = mg/(cos - µ_ssin)

Plugging this into the x-equation leads to:

v_max²/R = g(sin + µ_scos)/(cos - µ_µsin)

v_max² = (150 m)(9.8 m/s²)(sin18° + 0.3cos18°)/(cos18° - 0.3sin18°)

which gives

v_max = 32 m/s

F_N = mg/(cos + µ_s sin)

v_min²/R = g(sin - µ_scos)/(cos - µ_ssin)

Chapter 6

Problem 5)

The work done on the crate increases its kinetic energy:

W = K = ½mv² - ½mv₀²

= ½(66 kg)[(63 km/h)(10³ m/km)/(3600 s/h)]² - 0 = 1.01 x 10⁴ J

Another way to solve this would be to use W = Fx, and use kinematics to work out what x is.

Problem 6)

a) Because the workmen do not provide a force through a distance,     W = 0   (of course we are not considering the work done in pushing the piano to the window from wherever it was in the apartment!)

b) If the piano is slowly lowered, this means that the acceleration is very close to 0, so the tension has the same magnitude as the gravitational force:

W_T = - T x = - (mg)x = - (120 kg)(9.8 m/s²)(30 m) = - 3.53 x 10⁴ J

c) W_g = (mg)x = + 3.53 x 10⁴ J (equal and opposite to the work done by the tension)

Problem 45)

The work is done by two forces, the tension in the rope (provided by the movers) and gravity. Gravity is a conservative force, so the work done by gravity is independent of the path, and so it is the same for both paths. The net work is zero (since there is no change in kinetic energy!), and therefore the work done by the tension in the rope must also be the same for both paths:

W_net = 0 = W_T + W_g

W_T = - W_g = -mg(-H) = (54 kg)(9.8 m/s²)(4.0 m) = 2.1 x 10³ J

However, despite the fact that the work done by the movers is the same in both cases, they have to exert a smaller force (albeit over a larger distance) when they pull it up the incline, so they will likely consider that the easier way to go.

Problem 83)

a) The tension in the rope is always perpendicular to the direction of motion, thus it does no work: W_T = 0

b) Since the normal force (N) balances the puck's weight (mg), we have a frictional force

f_k = µ_kN = µ_kmg

The frictional force is opposite to the direction of motion, and therefore is tangential to the path. The work done by friction is then

W_f = µ_kmg(2r)

= -0.02(0.2 kg)(9.8 m/s²)(2)(0.8 m) = -0.20 J

The net work is equal to the change in kinetic energy, so

W = K

-0.20 J = K_f - (1/2)(0.2 kg)(10 m/s)²

which gives    K_f = 9.8 J