Force and Motion Problems: Solutions and Calculations

Posted on Dec 16, 2024 in Physics

Force and Motion: Solved Problems

Strength: solutions

1. Problem 1

   A body with a mass of 600 kg accelerates at 1.2 m/s². What force caused this acceleration?

   Data:

   • m = 600 kg
   • a = 1.2 m/s²

   Solution:

   F = ma = 600 kg * 1.2 m/s² = 720 N

2. Problem 2

   What mass must a body have for a force of 588 N to accelerate it at 9.8 m/s²?

   Data:

   • F = 588 N
   • a = 9.8 m/s²

   Solution:

   m = F / a = 588 N / 9.8 m/s² = 60 kg

3. Problem 3

   A 250 kg body is subjected to two opposing forces: 5880 N to the right and 5000 N to the left. What is the body’s acceleration?

   Data:

   • m = 250 kg
   • F_right = 5880 N
   • F_left = 5000 N

   Solution:

   a = (F_right – F_left) / m = (5880 N – 5000 N) / 250 kg = 3.52 m/s²

4. Problem 4

   Using the previous problem, if the forces act for one minute, how far will the body travel, and what will its final speed be?

   Data:

   • a = 3.52 m/s²
   • t = 1 min = 60 s
   • v_i = 0 m/s

   Solution:

   d = v_it + (at²) / 2 = 0 m/s * 60 s + 3.52 m/s² * (60 s)² / 2 = 6336 m

   v_f = v_i + at = 0 m/s + 3.52 m/s² * 60 s = 211.2 m/s

5. Problem 5

   A 100 kg body travels 1 km in 10 s from rest with constant acceleration. What force caused this?

   Data:

   • m = 100 kg
   • d = 1 km = 1000 m
   • t = 10 s
   • v_i = 0 m/s

   Solution:

   a = 2(d – v_it) / t² = 2(1000 m – 0 m/s * 10 s) / (10 s)² = 20 m/s²

   F = ma = 100 kg * 20 m/s² = 2000 N

F_right

F_left

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6. Problem 6

   A 3200 kg elevator falls with an acceleration of 1 m/s². Find the tension in the cable.

   Note: A force diagram is essential for this problem.

   Data:

   • m = 3200 kg
   • a = 1 m/s²
   • g = 9.8 m/s²

   Solution:

   T = mg – ma = m(g – a) = 3200 kg * (9.8 m/s² – 1 m/s²) = 28160 N

7. Problem 7

   A 2 kg body hangs from a cable. Find the tension if the acceleration is a) 5 m/s² upward, b) 5 m/s² downward.

   Data:

   • m = 2 kg
   • a) a = 5 m/s² upwards
   • b) a = 5 m/s² downwards

   Solution:

   a) T = ma + mg = m(a + g) = 2 kg * (5 m/s² + 9.8 m/s²) = 29.6 N

   b) T = mg – ma = m(g – a) = 2 kg * (9.8 m/s² – 5 m/s²) = 9.6 N

8. Problem 8

   Calculate the maximum acceleration a 90 kg man can have sliding down a rope that can only hold 735 N.

   Data:

   • m = 90 kg
   • T = 735 N

   Solution:

   a = (mg – T) / m = (90 kg * 9.8 m/s² – 735 N) / 90 kg = 1.633 m/s²

to

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9. Problem 9

   Two masses, 7 kg and 9 kg, hang from a string over a pulley. Assuming no friction, calculate the acceleration and tension.

   This is an Atwood machine.

   Data:

   • m₁ = 7 kg
   • m₂ = 9 kg

   Solution:

   a = (m₂ – m₁)g / (m₁ + m₂) = (9 kg – 7 kg) * 9.8 m/s² / (7 kg + 9 kg) = 1.225 m/s²

   T = m₁(a + g) = 7 kg * (1.225 m/s² + 9.8 m/s²) = 77.175 N

10. Problem 10

    A 50 kg block rests on a level surface. The minimum horizontal force to start movement is 147 N, and to keep it moving is 98 N. a) Calculate the kinetic friction coefficient, b) what is the friction force when a 49 N force is applied?

    Data:

    • m = 50 kg
    • F_{min_start} = 147 N
    • F_{min_keep} = 98 N

    Solution:

    a) μ_k = F / mg = 98 N / (50 kg * 9.8 m/s²) = 0.2

    b) The block will not move, so the static friction force is 49 N.

Note: Do not confuse the normal force (N) with the unit of force, the newton (N).

Consider that the static friction force is equal to the applied force until the minimum force to move is reached.

11. Problem 11

    A 50 kg block on a horizontal surface is pushed with 196 N for 3 s. The friction coefficient is 0.25. Find the final velocity.

    Data:

    • m = 50 kg
    • F = 196 N
    • t = 3 s
    • μ = 0.25

    Solution:

    a = (F – μmg) / m = (196 N – 0.25 * 50 kg * 9.8 m/s²) / 50 kg = 1.47 m/s²

    v_f = v_i + at = 0 m/s + 1.47 m/s² * 3 s = 4.41 m/s

12. Problem 12

    A 25 kg block on a table is connected by a cable over a pulley to a 20 kg hanging mass. What force must be applied to the 25 kg block to accelerate the 20 kg mass at 1 m/s², given a friction coefficient of 0.2?

    Data:

    • m₁ = 25 kg
    • m₂ = 20 kg
    • a = 1 m/s²
    • μ = 0.2

    Solution:

    F = (m₁ + m₂)a + (μm₁ + m₂)g = (25 kg + 20 kg) * 1 m/s² + (0.2 * 25 kg + 20 kg) * 9.8 m/s² = 302.25 N

mg NF fv 25 kg 20 kg FT TNM 1 gm 2 g aa f

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13. Problem 13

    A 100 kg body hangs from a rope. Calculate the acceleration when the tension is a) 125 N, b) 1200 N, c) 980 N.

    Data:

    • m = 100 kg
    • a) T = 125 N
    • b) T = 1200 N
    • c) T = 980 N

    Solution:

    a) a = (T – mg) / m = (125 N – 100 kg * 9.8 m/s²) / 100 kg = -8.55 m/s² (falls)

    b) a = (1200 N – 100 kg * 9.8 m/s²) / 100 kg = 2.2 m/s² (rises)

    c) a = (980 N – 100 kg * 9.8 m/s²) / 100 kg = 0 m/s² (constant velocity or at rest)

14. Problem 14

    A mine lift weighing 7840 N starts up with an acceleration of 6 m/s². Find the tension in the cable.

    Data:

    • W = mg = 7840 N
    • a = 6 m/s²

    Solution:

    m = 7840 N / 9.8 m/s² = 800 kg

    T = ma + mg = 800 kg * 6 m/s² + 7840 N = 12640 N

15. Problem 15

    A 1500 kg body hangs from a rope and descends at 4 m/s when it starts to stop. It stops in 3 m. Calculate the tension.

    Data:

    • m = 1500 kg
    • v_i = 4 m/s
    • v_f = 0 m/s
    • d = 3 m

    Solution:

    a = (v_f² – v_i²) / 2d = (0 m/s² – 4 m/s²) / (2 * 3 m) = -2.67 m/s²

    T = m(g – a) = 1500 kg * (9.8 m/s² – (-2.67 m/s²)) = 18700 N

mg N T T T mg ma mg m

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16. Problem 16

    A lift has a mass of 1200 kg. Find the tension when a) it accelerates at 1 m/s², b) it falls at 1 m/s².

    Data:

    • m = 1200 kg
    • a) a = 1 m/s² upwards
    • b) a = 1 m/s² downwards

    Solution:

    a) T = m(a + g) = 1200 kg * (1 m/s² + 9.8 m/s²) = 12960 N

    b) T = m(g – a) = 1200 kg * (9.8 m/s² – 1 m/s²) = 10560 N

17. Problem 17

    An 80 kg man is in an elevator descending at 1 m/s². Calculate the force exerted on the man by the lift. Repeat for ascending.

    Data:

    • m = 80 kg
    • a = 1 m/s² downwards

    Solution:

    Descending: T = m(g – a) = 80 kg * (9.8 m/s² – 1 m/s²) = 704 N

    Ascending: T = m(a + g) = 80 kg * (1 m/s² + 9.8 m/s²) = 864 N

18. Problem 18

    Two loads of 2 kg and 6 kg hang from a rope over a frictionless pulley. Calculate the acceleration and tension.

    This is similar to problem 9.

    Results:

    a = 4.9 m/s²

    T = 29.4 N

19. Problem 19

    An elevator starts upward with constant acceleration, rising 1 m in 0.8 s. A 3 kg package hangs by a thread. Find the tension.

    Data:

    • t = 0.8 s
    • d = 1 m
    • v_i = 0 m/s
    • m = 3 kg

    Solution:

    a = 2(d – v_it) / t² = 2(1 m – 0 m/s * 0.8 s) / (0.8 s)² = 3.125 m/s²

    T = m(a + g) = 3 kg * (3.125 m/s² + 9.8 m/s²) = 38.775 N

mg NT

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20. Problem 20

    A 70 kg skydiver falls freely for 5 s, then opens a parachute. It takes 0.8 s to fully open, increasing the velocity to 12 m/s. Calculate the average force on the parachute strings.

    Data:

    • m = 70 kg
    • v_i = 0 m/s
    • t_fall = 5 s
    • t_open = 0.8 s
    • v_f = 12 m/s

    Solution:

    v_fall = gt = 9.8 m/s² * 5 s = 49 m/s

    a = (v_f – v_fall) / t_open = (12 m/s – 49 m/s) / 0.8 s = -46.25 m/s²

    T = m(a + g) = 70 kg * (46.25 m/s² + 9.8 m/s²) = 3923.5 N

21. Problem 21

    A 50 kg block on a horizontal surface is pulled by a chord over a frictionless pulley to a 12 kg hanging mass. The friction coefficient is 0.2. Calculate the distance traveled in 10 s.

    Data:

    • m₁ = 50 kg
    • m₂ = 12 kg
    • μ = 0.2
    • t = 10 s
    • v_i = 0 m/s

    Solution:

    a = g(m₂ – μm₁) / (m₁ + m₂) = 9.8 m/s² * (12 kg – 0.2 * 50 kg) / (50 kg + 12 kg) = 0.316 m/s²

    d = v_it + (at²) / 2 = 0 m/s * 10 s + 0.316 m/s² * (10 s)² / 2 = 15.8 m

T mg to 50 kg

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