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Step-by-step solution for: Solved WORK AND THE WORK-ENERGY THEOREM Mech HW-47 Name 1. A ...
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---
## Problem Summary
We have two blocks, A and B, being pushed together by a hand along a frictionless table for a distance d.
Given:
- \( m_A > m_B \) (mass of block A is greater than mass of block B)
- The blocks move together — meaning they accelerate at the same rate.
- We are to analyze forces, work, kinetic energy, etc.
---
## Part a: Free-body diagrams and work between blocks
Since the table is frictionless, we only consider horizontal forces (and vertical forces cancel out: normal force = weight).
#### Block A:
- Force from hand → pushing to the right: \( F_{\text{hand}} \)
- Force from Block B → pushing to the left (Newton’s 3rd Law reaction): \( F_{B \to A} \)
- So net force on A: \( F_{\text{net},A} = F_{\text{hand}} - F_{B \to A} \)
#### Block B:
- Force from Block A → pushing to the right: \( F_{A \to B} \)
- (No other horizontal forces — no friction, no hand directly on B)
- So net force on B: \( F_{\text{net},B} = F_{A \to B} \)
> Note: By Newton’s 3rd Law, \( F_{A \to B} = -F_{B \to A} \), so magnitudes are equal, directions opposite.
---
✔ Yes.
Block A exerts a force \( F_{A \to B} \) to the right on block B, and block B moves to the right through distance \( d \). Since force and displacement are in the same direction, positive work is done by A on B.
Work done by A on B:
\[ W_{A \to B} = F_{A \to B} \cdot d \]
---
✔ Yes.
Block B exerts a force \( F_{B \to A} \) to the left on block A, while block A moves to the right. So force and displacement are in opposite directions → negative work is done by B on A.
Work done by B on A:
\[ W_{B \to A} = -F_{B \to A} \cdot d \]
But since \( F_{A \to B} = F_{B \to A} \) in magnitude, we have:
\[ |W_{A \to B}| = |W_{B \to A}| \]
but signs are opposite.
---
- Magnitude: Equal
- Sign: Opposite
- Explanation: This is consistent with Newton’s 3rd Law — action-reaction pairs have equal magnitude and opposite direction. Since both blocks move the same distance, the work magnitudes are equal, but signs differ because one force is in the direction of motion (A→B), the other is opposite (B→A).
---
## Part b: Net force on A vs. net force on B
From part a:
- \( F_{\text{net},A} = F_{\text{hand}} - F_{B \to A} \)
- \( F_{\text{net},B} = F_{A \to B} = F_{B \to A} \) (in magnitude)
But since both blocks move together, they have the same acceleration, \( a \).
So using Newton’s 2nd Law:
\[ F_{\text{net},A} = m_A \cdot a \]
\[ F_{\text{net},B} = m_B \cdot a \]
Since \( m_A > m_B \), then:
> ✔ Net force on block A is greater than net force on block B.
---
Recall: Work-Energy Theorem: Net work done on an object equals its change in kinetic energy.
Also, Work = Force × Displacement × cosθ
Since both blocks move the same distance \( d \), and the net force on each is constant and in the direction of motion:
\[ W_{\text{net},A} = F_{\text{net},A} \cdot d = m_A a d \]
\[ W_{\text{net},B} = F_{\text{net},B} \cdot d = m_B a d \]
So:
> ✔ Net work done on block A is greater than net work done on block B, because \( m_A > m_B \) and \( a \) and \( d \) are the same.
This makes sense — more massive object needs more work to achieve the same acceleration over same distance.
---
## Part c: Final kinetic energies comparison
Use Work-Energy Theorem:
\[ \Delta KE = W_{\text{net}} \]
Assume both start from rest? Not specified — but in part d, initial speed is given as 2 m/s, so let’s assume for now that both blocks start with the same initial velocity (since they’re pushed together).
Actually, part d gives initial speed = 2 m/s for both, so we’ll use that.
But even without numbers, we can reason:
Since both blocks start with same initial velocity (same initial KE per unit mass), and net work done on A > net work done on B, then:
> ✔ Final KE of block A > Final KE of block B
Because:
\[ KE_{\text{final}} = KE_{\text{initial}} + W_{\text{net}} \]
Initial KE:
\[ KE_{i,A} = \frac{1}{2} m_A v^2 \]
\[ KE_{i,B} = \frac{1}{2} m_B v^2 \]
Since \( m_A > m_B \), even initial KE of A is larger.
And net work on A is also larger → so final KE of A is definitely larger.
---
## Part d: Calculate final kinetic energy of each block
Given:
- Initial speed: \( v_i = 2 \, \text{m/s} \)
- Work done on block A by hand: \( W_{\text{hand on A}} = 10 \, \text{J} \)
- \( m_A = 4 \, \text{kg}, \, m_B = 1 \, \text{kg} \)
We need to find final KE of each block.
But note: the hand only pushes block A. However, block A pushes block B. So we must be careful — the 10 J is the work done *by the hand on block A*. But block A also does work on block B, so not all 10 J goes into increasing KE of A alone.
Actually, the 10 J is the total external work done on the system (since table is frictionless, no other external forces horizontally). So this 10 J should equal the total change in kinetic energy of the system.
Wait — let’s think carefully.
The hand applies a force to block A, and moves it a distance d. The work done by the hand is 10 J. Since the hand is external, and no other external horizontal forces, this 10 J is the total net work done on the entire system (A+B).
So:
> ✔ Total work done on system = 10 J = ΔKE_total
But we need individual KEs.
We can use the fact that both blocks have the same acceleration and same displacement, so we can find their individual net works from earlier.
Recall from part b:
Net force on A: \( F_{\text{net},A} = m_A a \)
Net force on B: \( F_{\text{net},B} = m_B a \)
Total net force on system: \( F_{\text{net,total}} = (m_A + m_B) a \)
Work done by hand = total net work on system = \( F_{\text{net,total}} \cdot d = (m_A + m_B) a d = 10 \, \text{J} \)
Now, net work on A: \( W_{\text{net},A} = m_A a d \)
Net work on B: \( W_{\text{net},B} = m_B a d \)
So the ratio of net work on A to net work on B is \( m_A : m_B = 4 : 1 \)
Therefore:
Total net work = \( W_{\text{net},A} + W_{\text{net},B} = 10 \, \text{J} \)
Let \( W_{\text{net},B} = x \), then \( W_{\text{net},A} = 4x \)
So:
\[ 4x + x = 10 \Rightarrow 5x = 10 \Rightarrow x = 2 \]
Thus:
- Net work on A = 8 J
- Net work on B = 2 J
Now compute initial KE:
Initial KE of A:
\[ KE_{i,A} = \frac{1}{2} m_A v_i^2 = \frac{1}{2} \cdot 4 \cdot (2)^2 = 2 \cdot 4 = 8 \, \text{J} \]
Initial KE of B:
\[ KE_{i,B} = \frac{1}{2} m_B v_i^2 = \frac{1}{2} \cdot 1 \cdot 4 = 2 \, \text{J} \]
Final KE:
- For A: \( KE_{f,A} = KE_{i,A} + W_{\text{net},A} = 8 + 8 = 16 \, \text{J} \)
- For B: \( KE_{f,B} = KE_{i,B} + W_{\text{net},B} = 2 + 2 = 4 \, \text{J} \)
✔ Final Answer:
- Final KE of block A = 16 J
- Final KE of block B = 4 J
---
## ✔ Summary of Answers
- Block A: Hand force → right, B-on-A force ← left
- Block B: A-on-B force → right
- Block A does positive work on B.
- Block B does negative work on A.
- Magnitudes equal, signs opposite — due to Newton’s 3rd Law and same displacement.
→ Net work on A > net work on B (same displacement).
- Block A: 16 J
- Block B: 4 J
---
Let me know if you’d like diagrams sketched or further clarification!
---
## Problem Summary
We have two blocks, A and B, being pushed together by a hand along a frictionless table for a distance d.
Given:
- \( m_A > m_B \) (mass of block A is greater than mass of block B)
- The blocks move together — meaning they accelerate at the same rate.
- We are to analyze forces, work, kinetic energy, etc.
---
## Part a: Free-body diagrams and work between blocks
Free-body Diagrams
Since the table is frictionless, we only consider horizontal forces (and vertical forces cancel out: normal force = weight).
#### Block A:
- Force from hand → pushing to the right: \( F_{\text{hand}} \)
- Force from Block B → pushing to the left (Newton’s 3rd Law reaction): \( F_{B \to A} \)
- So net force on A: \( F_{\text{net},A} = F_{\text{hand}} - F_{B \to A} \)
#### Block B:
- Force from Block A → pushing to the right: \( F_{A \to B} \)
- (No other horizontal forces — no friction, no hand directly on B)
- So net force on B: \( F_{\text{net},B} = F_{A \to B} \)
> Note: By Newton’s 3rd Law, \( F_{A \to B} = -F_{B \to A} \), so magnitudes are equal, directions opposite.
---
Does block A do work on block B?
✔ Yes.
Block A exerts a force \( F_{A \to B} \) to the right on block B, and block B moves to the right through distance \( d \). Since force and displacement are in the same direction, positive work is done by A on B.
Work done by A on B:
\[ W_{A \to B} = F_{A \to B} \cdot d \]
---
Does block B do work on block A?
✔ Yes.
Block B exerts a force \( F_{B \to A} \) to the left on block A, while block A moves to the right. So force and displacement are in opposite directions → negative work is done by B on A.
Work done by B on A:
\[ W_{B \to A} = -F_{B \to A} \cdot d \]
But since \( F_{A \to B} = F_{B \to A} \) in magnitude, we have:
\[ |W_{A \to B}| = |W_{B \to A}| \]
but signs are opposite.
---
Comparison:
- Magnitude: Equal
- Sign: Opposite
- Explanation: This is consistent with Newton’s 3rd Law — action-reaction pairs have equal magnitude and opposite direction. Since both blocks move the same distance, the work magnitudes are equal, but signs differ because one force is in the direction of motion (A→B), the other is opposite (B→A).
---
## Part b: Net force on A vs. net force on B
From part a:
- \( F_{\text{net},A} = F_{\text{hand}} - F_{B \to A} \)
- \( F_{\text{net},B} = F_{A \to B} = F_{B \to A} \) (in magnitude)
But since both blocks move together, they have the same acceleration, \( a \).
So using Newton’s 2nd Law:
\[ F_{\text{net},A} = m_A \cdot a \]
\[ F_{\text{net},B} = m_B \cdot a \]
Since \( m_A > m_B \), then:
> ✔ Net force on block A is greater than net force on block B.
---
What does this suggest about net work done on each block?
Recall: Work-Energy Theorem: Net work done on an object equals its change in kinetic energy.
Also, Work = Force × Displacement × cosθ
Since both blocks move the same distance \( d \), and the net force on each is constant and in the direction of motion:
\[ W_{\text{net},A} = F_{\text{net},A} \cdot d = m_A a d \]
\[ W_{\text{net},B} = F_{\text{net},B} \cdot d = m_B a d \]
So:
> ✔ Net work done on block A is greater than net work done on block B, because \( m_A > m_B \) and \( a \) and \( d \) are the same.
This makes sense — more massive object needs more work to achieve the same acceleration over same distance.
---
## Part c: Final kinetic energies comparison
Use Work-Energy Theorem:
\[ \Delta KE = W_{\text{net}} \]
Assume both start from rest? Not specified — but in part d, initial speed is given as 2 m/s, so let’s assume for now that both blocks start with the same initial velocity (since they’re pushed together).
Actually, part d gives initial speed = 2 m/s for both, so we’ll use that.
But even without numbers, we can reason:
Since both blocks start with same initial velocity (same initial KE per unit mass), and net work done on A > net work done on B, then:
> ✔ Final KE of block A > Final KE of block B
Because:
\[ KE_{\text{final}} = KE_{\text{initial}} + W_{\text{net}} \]
Initial KE:
\[ KE_{i,A} = \frac{1}{2} m_A v^2 \]
\[ KE_{i,B} = \frac{1}{2} m_B v^2 \]
Since \( m_A > m_B \), even initial KE of A is larger.
And net work on A is also larger → so final KE of A is definitely larger.
---
## Part d: Calculate final kinetic energy of each block
Given:
- Initial speed: \( v_i = 2 \, \text{m/s} \)
- Work done on block A by hand: \( W_{\text{hand on A}} = 10 \, \text{J} \)
- \( m_A = 4 \, \text{kg}, \, m_B = 1 \, \text{kg} \)
We need to find final KE of each block.
But note: the hand only pushes block A. However, block A pushes block B. So we must be careful — the 10 J is the work done *by the hand on block A*. But block A also does work on block B, so not all 10 J goes into increasing KE of A alone.
Actually, the 10 J is the total external work done on the system (since table is frictionless, no other external forces horizontally). So this 10 J should equal the total change in kinetic energy of the system.
Wait — let’s think carefully.
The hand applies a force to block A, and moves it a distance d. The work done by the hand is 10 J. Since the hand is external, and no other external horizontal forces, this 10 J is the total net work done on the entire system (A+B).
So:
> ✔ Total work done on system = 10 J = ΔKE_total
But we need individual KEs.
We can use the fact that both blocks have the same acceleration and same displacement, so we can find their individual net works from earlier.
Recall from part b:
Net force on A: \( F_{\text{net},A} = m_A a \)
Net force on B: \( F_{\text{net},B} = m_B a \)
Total net force on system: \( F_{\text{net,total}} = (m_A + m_B) a \)
Work done by hand = total net work on system = \( F_{\text{net,total}} \cdot d = (m_A + m_B) a d = 10 \, \text{J} \)
Now, net work on A: \( W_{\text{net},A} = m_A a d \)
Net work on B: \( W_{\text{net},B} = m_B a d \)
So the ratio of net work on A to net work on B is \( m_A : m_B = 4 : 1 \)
Therefore:
Total net work = \( W_{\text{net},A} + W_{\text{net},B} = 10 \, \text{J} \)
Let \( W_{\text{net},B} = x \), then \( W_{\text{net},A} = 4x \)
So:
\[ 4x + x = 10 \Rightarrow 5x = 10 \Rightarrow x = 2 \]
Thus:
- Net work on A = 8 J
- Net work on B = 2 J
Now compute initial KE:
Initial KE of A:
\[ KE_{i,A} = \frac{1}{2} m_A v_i^2 = \frac{1}{2} \cdot 4 \cdot (2)^2 = 2 \cdot 4 = 8 \, \text{J} \]
Initial KE of B:
\[ KE_{i,B} = \frac{1}{2} m_B v_i^2 = \frac{1}{2} \cdot 1 \cdot 4 = 2 \, \text{J} \]
Final KE:
- For A: \( KE_{f,A} = KE_{i,A} + W_{\text{net},A} = 8 + 8 = 16 \, \text{J} \)
- For B: \( KE_{f,B} = KE_{i,B} + W_{\text{net},B} = 2 + 2 = 4 \, \text{J} \)
✔ Final Answer:
- Final KE of block A = 16 J
- Final KE of block B = 4 J
---
## ✔ Summary of Answers
a. Free-body diagrams:
- Block A: Hand force → right, B-on-A force ← left
- Block B: A-on-B force → right
- Block A does positive work on B.
- Block B does negative work on A.
- Magnitudes equal, signs opposite — due to Newton’s 3rd Law and same displacement.
b. Net force on A > net force on B (since \( m_A > m_B \), same acceleration).
→ Net work on A > net work on B (same displacement).
c. Final KE of A > final KE of B — because both start with same speed, but A has more mass and more net work done on it.
d. Final KE:
- Block A: 16 J
- Block B: 4 J
---
Let me know if you’d like diagrams sketched or further clarification!
Parent Tip: Review the logic above to help your child master the concept of work energy theorem worksheet.
