Problems

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

A box of mass 5.0 kg is pulled vertically upwards by a force of 68 N applied to a rope attached to the box. Find a) the acceleration of the box and b) the vertical velocity of the box after 2 seconds.

Solution:

 

Figure 4.1: Problem 4.1

a)

From the 2nd Law:

ma = T - mg

$$\Rightarrow {\frac{T}{m}}$$ =

$${\frac{68 \;{\:\rm N}}{5 \;{\:\rm kg}}}$$ = (5)

b)

Since a is constant

v = v₀ + at

= 0 + 3.8(2) = 7.6  m/s. (6)

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

A hockey puck of mass .5 kg travelling at 10 m/s slows to 2.0 m/s over a distance of 80 m. Find a) the frictional force acting on the puck and b) the coefficient of kinetic friction between the puck and the surface.

Solution:

 

Figure 4.2: Problem 4.2

a)

First we find the acceleration of the puck from the kinematic equations of motion. We have, v₀ = 10  m/s , v = 2  m/s and x = 80  m . The third equation of motion gives,

v ² = v₀² + 2ax

$$\rightarrow {\frac{v^2-v_0^2}{2x}} {\frac{4-100}{2(80)}}$$ = (7)

From the Second Law:
In the x-direction,

f_k = ma = .5(- 0.6) = - 0.3  N. (8)

b)

Use f_k = - $\mu_{k}^{}$N . From the y-component of the 2nd Law: N - mg = 0 . Combining,

$$\mu_{k}^{}$$ = ma

$$\rightarrow \mu_{k}^{}$$ = - a/g

= 0.061. (9)

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

A student of mass 50 kg tests Newton's laws by standing on a bathroom scale in an elevator. Assume that the scale reads in newtons. Find the scale reading when the elevator is a) accelerating upward at .5 m/s ², b) going up at a constant speed of 3.0 m/s and c) going up but decelerating at 1.0 m/s ².

Solution:

 

Figure 4.3: Problem 4.3

From the 2nd Law:

F_s - mg = ma

$$\rightarrow$$ = m(g + a). (10)

This gives:

a)

F_s = 50(9.8 + 0.5) = 515  N

b)

F_s = 50(9.8 + 0) = 490  N

c)

F_s = 50(9.8 - 1.0) = 440  N

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

A wooden plank is raised at one end to an angle of 30 ^o . A 2.0 kg box is placed on the incline 1.0 m from the lower end and given a slight tap to overcome static friction. The coefficient of kinetic friction between the box and the plank is $\mu_{k}^{}$ = 0.20 . Find a) the rate of acceleration of the box and b) the speed of the box at the bottom. Assume that the initial speed of the box is zero.

Solution:

 

Figure 4.4: Problem 4.4

a)

Find the components of the weight of the object:

$$\theta$$ w_x =

$$\theta$$ w_y =

Write out the two components of Newton's 2nd Law:

$$\theta$$ x

$$\theta$$ y (11)

Using f_k = $\mu_{k}^{}$N we get,

$$\mu_{k}^{} \theta \theta$$ ma =

$$\rightarrow \theta \mu_{k}^{} \theta$$ =

= 9.8(sin 30 - 0.2cos 30) = 3.20 m/s ² (12)

b)

Since a is constant and v ² = v₀² + 2ax . With x = 1  m , v₀ = 0 we have

$$\sqrt{2(3.2)(1)}$$

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

A 10 kg box is attached to a 7 kg box which rests on a 30 ^o incline. The coefficient of kinetic friction between each box and the surface is $\mu_{k}^{}$ = .1 . Find a) the rate of acceleration of the system and b) the tension in the rope.

Solution:

 

Figure 4.5: Problem 4.5

We apply the 2nd law separately to each box.
For the 10 kg box:
y direction:

N₂ - m₂g =

N₂ = m₂g,

x direction:

T - f_{k 2} = m₂a

$$\mu_{k}^{}$$ = m₂a

$$\rightarrow \mu_{k}^{}$$ = m₂a. (13)

For the 7 kg box:
y direction:

$$\theta$$

x direction:

$$\theta$$ = m₁a

$$\theta \mu_{k}^{}$$ = m₁a

$$\rightarrow \theta \mu_{k}^{} \theta$$ = m₁a. (14)

We have a system of two equations and two unknowns: a and T . We can solve as follows.

a)

From equation (4.1), T = m₂a + $\mu_{k}^{}$m₂g . Substituting into equation (4.2) gives,

$$\theta \mu_{k}^{} \theta \mu_{k}^{}$$ m₁a =

$$\rightarrow \theta \mu_{k}^{} \mu_{k}^{} \theta$$ = (15)

$$\rightarrow {\textstyle\frac{1}{m_1+m_2}} \theta \mu_{k}^{} \mu_{k}^{} \theta \mu_{k}^{}$$ =

$${\textstyle\frac{1}{17}}$$ = (16)

b)

Then substituting into the first equation gives,

$$\mu_{k}^{}$$ T =

= 10(9.8(.1) + 1.1) = 20.8  N. (17)