PEP 294: Solutions

XII. Linear Kinetics

Useful Equations

(1)    Newton’s law of acceleration:     F = ma

(2)    Frictional force:    F_f = mN

(3)    Weight:    F = mg   (where g = 9.81 m/s²)

(4)    Momentum:    M = mv

(5)    Impulse:    I = Ft

(6)    Work:    W = Fd (cos q)

(7)    Power:  

(8)    Potential energy:    PE = mgh

(9)    Kinetic energy:   

Introductory Problems

1. Ball:  m = 2.5 kg, a = 40 m/s²  =>  force

Use Eq. 1 to solve this problem:

F = ma = (2.5)(40) = 100 N

2. High jumper:  weight = 712 N, F = 3 kN = 3,000 N  =>  reaction force?

According to the Newton’s Third Law of Motion, the reaction force is the same to the force in magnitude but in opposite direction:

F₁₂ = - F₂₁

Therefore, the magnitude of the reaction force is 3,000 N as well.

3. Factors affecting friction:

From Eq. 2, (a) normal reaction force acting perpendicular to the surface of contact between two contacting objects (N), and (b) the friction coefficient (m). The friction coefficient is a function of the condition of the contacting surfaces such as roughness, etc.

 

4. Basketball shoe:  m_s = 0.56, N = 350 N  =>  F causing shoe to slide

Normal reaction force (N) = weight of the basketball player

From Eq. 2:

F_f = m_s N = (0.56)(350) = 196 N

The basketball shoe slides when the horizontal force is larger than 196 N.

5. Tackling sled: weight = 670 N, m_s = 0.73, m_k = 0.68

Normal reaction force (N) = weight of the sled = 670 N

From Eq. 2:

Force required to start the sled (max static friction):

F_f(s) = m_s N = (0.73)(670) = 489.1 N

Force required to keep the sled in motion (kinetic friction):

F_f(k) = m_k N = (0.68)(670) = 455.6 N

For coach + sled:

m_coach = 100 kg  =>   weigh = (100)(9.81) = 981 N (Eq. 3)

N = weight of sled + weight of coach = 670 + 981 = 1,651 N

F_f(s) = m_sN = (0.73)(1651) = 1205.2 N

F_f(k) = m_kN = (0.68)(1651) = 1122.7 N

6. Collision -- Lineman A: m = 100 kg, v = 4 m/s; Lineman B: m = 90 kg, v = 4.5 m/s; Head-on collision  =>  velocity after impact

Head-on impact  => the direction of the velocities are opposite, so assume A is moving to the right.

The question implies a non-elastic impact.

Use Eq. 4 to compute momentum.

Before collision: A (100 kg), B (90 kg)

M_before = M_A + M_B = (100)(4) + (90)(-4.5) = 400 - 405 = - 5 kg m/s

After collision: A + B (190 kg)  =>  v

M_after = M_A+B = (190)(v)

Momentum conservation:

M_before = M_after  =>  - 5 = (190)(v)

v = (-5)/(190) = -0.03 m/s   (direction: left)

B pushes A at a speed of 0.03 m/s.

7. Use the approach used in problem 6. Do it by yourself. You may see this in the test.

9. 20 stairs (h_step = 20 cm), weight = 700 N, Dt = 1.25 s  =>  W, P, and DPE

Height of stairs = (0.2 m/step)(20 steps) = 4 m

Force required to overcome the resistance (weight) = 700 N (upward)

Combine d and cos q in Eq. 6, which is equal to the height of the stairs.

Work (Eq. 6):  W = Fd (cos q) = F (d cos q) = Fh = (700)(4) = 2,800 J

Power (Eq. 7):  P = W / Dt = (2800)/(1.25) = 2,240 W

Change in PE:  Same to work done  =>   2,800 J

10. Pitched ball:  m = 1 kg, v = 28 m/s

From Eq. 4:

Momentum of the ball:  M = m v = (1)(28) = 28 kg m/s

Impulse (I) = change in momentum (DM)   =>  in stopping the ball, I = DM = 28 kg m/s

I = 28 Ns

Since Dt = 0.5 s:

From Eq. 5:  I = (F_impuse)(Dt) =>  F_impuse = I / Dt = (28) / (0.5) = 56 N

Additional Problems

3. m_block = 2 kg, F = 7.5 N, resulting a = 3 m/s²   =>  F_f

[Hint] This problem is combination of the Newton’s Law of Acceleration and frictional force. See the diagram below. The net force (F - F_f) causes an acceleration of 3 m/s².

Eq. 1:  F_net = ma = (2)(3) = 6 N

F_net = F - F_f = 7.5 - F_f   =>  7.5 - F_f = 6

F_f = 7.5 - 6 = 1.5 N

6. Golf club:  r = 108 cm = 1.08 m, m = 0.73 kg, Dt = 0.5 s, a = 10 rad/s²  =>   M

This is combination of circular motion and linear momentum. Figure out the linear velocity of the club head first and compute the momentum. The question implies that the initial angular velocity of the club is 0 rad/s.

   

Angular velocity at impact:  w₂ = (10)(0.5) = 5 rad/s

Linear velocity of the club head:  v₂ = (r)(w₂) = (1.08)(5) = 5.4 m/s

Momentum of the head:  M₂ = (m)(v₂) = (0.73)(5.4) = 3.94 kg m/s

7. Ball:  weight = 6.5 N, initial v = 20 m/s, angle = 35°, projection height = 1.5 m

This problem is a combination of projectile motion, weight and kinetic energy.

The velocity of the ball at the height it was projected is the same to the initial velocity.

Mechanical work done in stopping the ball in motion is the same to the KE of the ball before the catch.

Mass of the ball, from Eq. 3:  m = (6.5)/(9.81) = 0.66 kg

Velocity at 1.5 m high = initial v = 20 m/s

From Eq. 9:  KE = (1/2)(m)(v²) = (0.5)(0.66)(20²) = 132.5 J

8. Vertical jump:  m = 50 kg, v = 2 m/s

Max. KE (at takeoff) = (0.5)(50)(2²) = 100 J

Max. PE, based on the ME conservation law (at the apex) = 100 J

Min. KE (at the apex) = 0 J

Elevation of jumper’s center of mass (CM):

Max. KE at takeoff = Max. PE at apex:

(0.5)(50)(2²) = (50)(9.81)(h)

h = 0.20 m

9. Apply the approach used in the previous problem. Do it by yourself.