Physics Problem Set: Work, Energy, and Force Calculations
Classified in Physics
Written on in 
English with a size of 24.88 KB
Physics Problems: Work, Energy, and Forces
This document presents a series of physics problems focusing on concepts of work, energy, and forces, followed by their detailed solutions. These exercises are designed to reinforce understanding of fundamental mechanics principles.
Problem Statements
Problem 1: Work as a Scalar or Vector?
Is work a vector or a scalar quantity?
Problem 2: Negative Work Scenarios
Can work be negative? If so, in what cases does it occur?
Problem 3: When is Energy Not Conserved?
In what cases is mechanical energy not conserved?
Problem 4: Work Done on a Box
A box weighing 900 [N] rests on the ground. Calculate the work required to move it at a constant speed:
a) Moving Horizontally Against Friction
4 [m] on the floor against a friction force of 180 [N]?
Result: W = 720 [J]
b) Lifting Vertically
4 [m] vertically upwards?
Result: W = 3600 [J]
Problem 5: Block Motion with Friction
A block of mass M = 4 [kg], initially at rest, is pulled by a constant force F1 = 15 [N], as shown in Figure 1. Find the speed of the block after it has moved 3 [m], if the surface has a coefficient of friction μ = 0.2.
Result: v = 3.12 [m/s]
Problem 6: Elastic Potential Energy and Work
A mass attached to a spring with an elasticity constant of 4 [N/m] is initially at -5 [cm] from its equilibrium position. A force then compresses the spring further to -8 [cm]. From this information, determine:
a) Initial Elastic Potential Energy
The elastic potential energy in the initial position.
Result: 0.005 [J]
b) Final Elastic Potential Energy
The elastic potential energy in the final position.
Result: 0.0128 [J]
c) Work Done by Elastic Force
The work done by the elastic force.
Result: -0.0078 [J]
Problem 7: Electric Cart and Inclined Plane
An electric motorized cart, capable of developing a force of 50 [N], must raise a 120 [kg] mass from the floor to a height of 10 [m].
a) Total Work Required
How much work must be done?
Result: W = 11,760 [J]
b) Inclined Plane Length
Since it cannot be lifted vertically, the owner decides to use an inclined plane. Neglecting friction, what must be the length of the incline?
Result: d = 235.2 [m]
Problem 8: Force Exerted by a Glove on a Baseball
A baseball (mass m = 140 [g]) travels at 35 [m/s] and moves 25 [cm] backward into a fielder's glove as it is caught. What was the average force exerted on the ball by the glove?
Result: F = 343 [N]
Detailed Solutions
Solution 1: Work as a Scalar
Work is a scalar quantity, as it is the result of a dot product between the force vector and the displacement vector. It only has magnitude, not direction.
Solution 2: Negative Work
Yes, work can be negative. This occurs when the force applied is in the opposite direction to the displacement. Common examples include:
- Work done by friction, which always opposes motion.
- Work done by a spring force when it acts against the direction of the object's movement.
- Work done by air resistance.
Solution 3: Non-Conservative Forces and Energy Conservation
Mechanical energy is not conserved when non-conservative forces perform work on the system. These forces, such as friction or air resistance, dissipate mechanical energy, converting it into other forms like heat or sound. The work done by non-conservative forces depends on the path taken, unlike conservative forces (e.g., gravity, spring force).
Solution 4: Work Calculations for a Box
a) Work Against Friction
To move the box horizontally against a friction force:
W = Ffriction ⋅ d
W = 180 [N] ⋅ 4 [m]
W = 720 [J]
b) Work to Lift Vertically
To lift the box vertically:
W = Weight ⋅ height
W = 900 [N] ⋅ 4 [m]
W = 3600 [J]
Solution 5: Block Speed with Friction
Given: M = 4 [kg], F1 = 15 [N], d = 3 [m], μ = 0.2, Vinitial = 0 [m/s]
1. Analyze Forces in Y-direction (Vertical):
ΣFy = N + F1 sin(θ) - mg = 0
N = mg - F1 sin(θ)
N = (4 [kg] ⋅ 9.8 [m/s²]) - (15 [N] ⋅ sin(32°))
N = 39.2 [N] - (15 ⋅ 0.5299)
N = 39.2 [N] - 7.9485 [N]
N ≈ 31.25 [N]
2. Calculate Friction Force:
Ffriction = μ ⋅ N
Ffriction = 0.2 ⋅ 31.25 [N]
Ffriction = 6.25 [N]
3. Calculate Net Force in X-direction (Horizontal):
ΣFx = F1 cos(θ) - Ffriction
ΣFx = (15 [N] ⋅ cos(32°)) - 6.25 [N]
ΣFx = (15 ⋅ 0.8480) - 6.25 [N]
ΣFx = 12.72 [N] - 6.25 [N]
ΣFx ≈ 6.47 [N]
4. Apply Work-Energy Theorem:
Wnet = ΔKE
ΣFx ⋅ d = ½mvf² - ½mvi²
Since Vinitial = 0:
6.47 [N] ⋅ 3 [m] = ½ ⋅ 4 [kg] ⋅ vf²
19.41 [J] = 2 ⋅ vf²
vf² = 19.41 / 2
vf² = 9.705
vf = √9.705
vf ≈ 3.12 [m/s]
Solution 6: Elastic Potential Energy and Work Calculations
Given: k = 4 [N/m], xinitial = -5 [cm] = -0.05 [m], xfinal = -8 [cm] = -0.08 [m]
a) Initial Elastic Potential Energy (Einitial)
Einitial = ½kxinitial²
Einitial = 0.5 ⋅ 4 [N/m] ⋅ (-0.05 [m])²
Einitial = 2 ⋅ 0.0025 [m²]
Einitial = 0.005 [J]
b) Final Elastic Potential Energy (Efinal)
Efinal = ½kxfinal²
Efinal = 0.5 ⋅ 4 [N/m] ⋅ (-0.08 [m])²
Efinal = 2 ⋅ 0.0064 [m²]
Efinal = 0.0128 [J]
c) Work Done by Elastic Force (Welastic)
The work done by the elastic force is the negative change in elastic potential energy:
Welastic = -(Efinal - Einitial) = Einitial - Efinal
Welastic = 0.005 [J] - 0.0128 [J]
Welastic = -0.0078 [J]
Solution 7: Electric Cart Work and Inclined Plane
Given: m = 120 [kg], h = 10 [m], Fcart = 50 [N]
a) Total Work Required to Lift
The work required to lift the mass vertically is equal to the change in gravitational potential energy:
W = mgh
W = 120 [kg] ⋅ 9.8 [m/s²] ⋅ 10 [m]
W = 11,760 [J]
b) Length of Inclined Plane
If friction is neglected, the work done by the cart along the incline must equal the work required to lift the mass vertically:
W = Fcart ⋅ d
11,760 [J] = 50 [N] ⋅ d
d = 11,760 [J] / 50 [N]
d = 235.2 [m]
Solution 8: Force Exerted by Glove on Baseball
Given: m = 140 [g] = 0.140 [kg], Vinitial = 35 [m/s], d = 25 [cm] = 0.25 [m], Vfinal = 0 [m/s] (ball stops)
Apply the Work-Energy Theorem:
Wnet = ΔKE
Faverage ⋅ d = ½mvf² - ½mvi²
Faverage ⋅ 0.25 [m] = ½ ⋅ 0.140 [kg] ⋅ (0 [m/s])² - ½ ⋅ 0.140 [kg] ⋅ (35 [m/s])²
Faverage ⋅ 0.25 = 0 - (0.070 ⋅ 1225)
Faverage ⋅ 0.25 = -85.75
Faverage = -85.75 / 0.25
Faverage = -343 [N]
The negative sign indicates that the force exerted by the glove is in the opposite direction to the ball's initial motion, which is expected as it slows the ball down. The magnitude of the average force is 343 [N].