Physics Problems: Motion, Force, and Energy Solutions

Posted on Jan 13, 2025 in Physics

Understanding Linear Motion with Constant Acceleration

1. A truck starts from rest and moves with an acceleration of 4 m/s2. Find its velocity and the distance traveled after 15 seconds.

  • Xf = Xi + Vit + at2/2
  • Vf = Vi + at

Since the truck starts from rest, Xi = 0 and Vi = 0.

  • Xf = at2/2
  • Vf = at
  • Xf = (4)(15)2/2
  • Vf = (4)(15)
  • Xf = 450 m
  • Vf = 60 m/s

Calculating Deceleration and Force in Linear Motion

2. A car of mass 1,200 kg, traveling at 25 m/s, stops completely in 6 seconds. What is the value of the acceleration and the force applied to make the car stop?

  • a = Vi/t
  • Xf = Xi + Vit + at2/2

Since the car stops completely, Vf = 0.

  • a = 25/6
  • Xf = Vit + at2/2
  • a = 4.16 m/s2
  • Xf = (25)(6) + (4.16)(6)2/2
  • Xf = 150 + 74.88
  • Xf = 224.88 m

Determining Acceleration from a Velocity-Time Graph

3. Calculate the acceleration of the body in the interval from 10 to 15 seconds.

  • a = (Vf – Vi) / (tf – ti)
  • a = (8 – 12) / (15 – 10)
  • a = -0.8 m/s2

Calculating Area Under a Velocity-Time Graph

4. Calculate the acceleration of the body in the interval of 20 to 25 seconds.

  • A = b * h
  • A = 5 * 4
  • A = 20 m

Solving Free Fall Problems

1. A bottle is dropped from a balloon and reaches the ground in 12 seconds. Find the velocity with which it hit the ground (Vf). (Free fall)

  • Vf = at + at
  • Vf = at
  • Vf = (-9.8)(12)
  • Vf = -117.6 m/s

Analyzing Vertical Projectile Motion

2. A ball is thrown vertically upward with an initial velocity of 39 m/s. Find the time it takes to reach the maximum height and the height.

  • Vf = Vi + at
  • Yf = Yi + Vit + at2/2

At maximum height, Vf = 0.

  • t = –Vi/a
  • Yf = Vit + at2/2
  • t = -39/-9.8 = 3.97 s
  • Yf = (39)(3.97) + (-9.8)(3.97)2/2
  • Yf = 154.83 – 77.22 = 77.61 m

Calculating Horizontal Range in Projectile Motion

3. A ball rolls on the edge of a table with an initial velocity of 26 m/s. If when it fell from the table it took 1.05 seconds to reach the ground, what is the range or distance traveled by the ball (Xf)? (Horizontal projectile motion)

  • Xf = Vi * t
  • Xf = (26)(1.05) = 27.3 m

Finding Maximum Height in Parabolic Projectile Motion

4. A projectile is launched with an initial velocity of 42 m/s at an angle of 25°. Find the maximum height that the projectile reaches if it takes 1.3 seconds to reach said height. (Parabolic projectile motion)

  • Yf = Yi + Vit + at2/2
  • Vix = Vi cos(angle)
  • Viy = Vi sin(angle)

Since we are looking for the maximum height, we use the vertical component of the initial velocity.

  • Yf = Vit + at2/2
  • Viy = 42 sin(25)
  • Viy = 17.74 m/s
  • Yf = (17.74)(1.3) + (-9.8)(1.3)2/2
  • Yf = 23.062 – 8.281 = 14.781 m

Angular Velocity Calculation

1. A Ferris wheel spins 540° and stops. If it took 30 seconds to complete that trajectory, what was its angular velocity?

First, convert degrees to radians: 540° * (π / 180°) = 3π rad ≈ 9.42 rad

  • ω = angle / t
  • ω = 9.42 / 30 = 0.314 rad/s

Tangential Velocity in Circular Motion

2. A carpenter uses a circular saw with a radius of 0.2 meters to cut wood. If the saw has an angular velocity of 18 rad/s, what is its tangential velocity?

  • v = ω * r
  • v = (18)(0.2) = 3.6 m/s

Centripetal Acceleration and Force

3. Calculate the centripetal acceleration and centripetal force of a 45 kg body that rotates in a radius of 6 meters with an angular velocity of 2.4 rad/s.

  • ac = ω2 * r
  • ac = (2.4)2(6) = 34.56 m/s2
  • Fc = m * ac
  • Fc = (45)(34.56) = 1,555.2 N

Torque Calculation for Loosening a Nut

4. A force of 64 N is applied to a wrench to loosen a nut. If the force is applied 0.03 m from the axis of rotation of the nut, determine the torque needed to loosen the nut.

  • τ = F * r
  • τ = (64)(0.03) = 1.92 N*m

Analyzing Forces and Motion with Friction

1. A wooden box with a mass of 15 kg is pushed by a force of 94 N on a horizontal surface. If the coefficient of kinetic friction is μk = 0.40, calculate the value of the normal force and acceleration and draw the free-body diagram of this analysis.

  • w = m * g
  • w = 15 * 9.8 = 147 N
  • FN = w = 147 N
  • f = μk (FN)
  • f = 0.40(147) = 58.8 N
  • Fnet = FA – f
  • Fnet = 94 – 58.8 = 35.2 N
  • Fnet = m * a
  • a = Fnet / m = 35.2 / 15 = 2.34 m/s2

Calculating Weight and Thrust Force in a Static System

2. A body of weight (w) is suspended in a structure. If the tension of the rope is 400 N with an angle of 40° with respect to the horizontal, calculate the weight of the body and the thrust force of the beam.

  • Tx = T cos(Θ)
  • Tx = 400 cos(40) = 306.41 N
  • Ty = T sin(Θ)
  • Ty = 400 sin(40) = 257.11 N
  • F – Tx = 0
  • F = Tx = 306.41 N
  • Ty – w = 0
  • Ty = w = 257.11 N