Question

in which of these examples is the greatest acceleration occurring?
a
50 n
50 n
b
75 n
50 n
c
10 n
50 n
d
100 n
100 n

Explanation:

Step1: Recall Newton's Second Law

Newton's second law states \( F_{net} = ma \), so \( a=\frac{F_{net}}{m} \). Assuming the mass \( m \) of the object (the cart) is the same in all cases, the acceleration is proportional to the net force. We calculate the net force for each option.

Step2: Calculate Net Force for Option A

For option A, the forces are \( 50\,\text{N} \) (right) and \( 50\,\text{N} \) (left). The net force \( F_{net,A}=50 - 50 = 0\,\text{N} \).

Step3: Calculate Net Force for Option B

For option B, the forces are \( 75\,\text{N} \) (right) and \( 50\,\text{N} \) (left). The net force \( F_{net,B}=75 - 50 = 25\,\text{N} \) (to the right).

Step4: Calculate Net Force for Option C

For option C, the forces are \( 10\,\text{N} \) (right) and \( 50\,\text{N} \) (left). The net force \( F_{net,C}=10 - 50=- 40\,\text{N} \) (to the left, magnitude \( 40\,\text{N} \)). Wait, correction: net force is the vector sum. So \( F_{net,C}=50 - 10 = 40\,\text{N} \) to the left? Wait no, direction: if left is positive, then \( 50 - 10 = 40\,\text{N} \) left. But magnitude is \( 40\,\text{N} \). Wait, no: force to the left is \( 50\,\text{N} \), right is \( 10\,\text{N} \). So \( F_{net}=50 - 10 = 40\,\text{N} \) left (magnitude 40 N). Wait, but let's re - check:

Wait, the formula is \( F_{net}=F_{right}-F_{left} \) if we take right as positive. So for A: \( 50 - 50 = 0 \); B: \( 75 - 50 = 25 \); C: \( 10 - 50=-40 \) (magnitude 40 N, direction left); D: \( 100 - 100 = 0 \).

Wait, but the mass is the same, so acceleration is proportional to the magnitude of net force. Wait, but in option C, the net force magnitude is 40 N, in B it's 25 N, A and D are 0. Wait, but maybe I made a mistake. Wait, the problem is about the greatest acceleration. Acceleration is a vector, but the magnitude of acceleration is \( |a|=\frac{|F_{net}|}{m} \). So we need to find the case with the largest \( |F_{net}| \).

Wait, let's recalculate:

Option A: \( F_{net}=50 - 50 = 0\,\text{N} \), \( |F_{net}| = 0 \)

Option B: \( F_{net}=75 - 50 = 25\,\text{N} \), \( |F_{net}| = 25 \)

Option C: \( F_{net}=10 - 50=-40\,\text{N} \), \( |F_{net}| = 40 \)

Option D: \( F_{net}=100 - 100 = 0\,\text{N} \), \( |F_{net}| = 0 \)

Wait, but this contradicts the initial thought. Wait, maybe the cart's mass is the same, so the acceleration magnitude is \( \frac{|F_{net}|}{m} \). So the larger the \( |F_{net}| \), the larger the acceleration. So option C has \( |F_{net}| = 40\,\text{N} \), option B has 25 N, so C has larger net force? But wait, maybe I misread the forces. Let's check the arrows again. In option C, the right force is 10 N, left is 50 N. So net force is 50 - 10 = 40 N to the left. In option B, right is 75 N, left is 50 N, net force 25 N to the right. So magnitude of net force in C is 40 N, in B is 25 N. So acceleration in C is \( \frac{40}{m} \), in B is \( \frac{25}{m} \), A and D are 0. But wait, maybe the mass is not the same? No, the problem is about the same cart, so mass is same. Wait, but maybe I made a mistake in the direction of the force. Wait, maybe the left force in C is 10 N and right is 50 N? No, the diagram shows: in C, left arrow (blue) is 50 N, right arrow (red) is 10 N. So net force is 50 - 10 = 40 N left.

Wait, but the answer might be different. Wait, maybe the question is about the magnitude of acceleration. So the largest magnitude of net force will give the largest acceleration. So between B (25 N) and C (40 N), C has larger net force. But wait, let's check again.

Wait, no, maybe I messed up the subtraction. Let's define the positive direction as the direction of the larger force. For option B: forces are 75 N right and 50 N left. Net force is 75 - 50 = 25 N right. For option C: 50 N left and 10 N right. Net force is 50 - 10 = 40 N left. So the magnitude of net force in C is 40 N, in B is 25 N. So acceleration in C is \( \frac{40}{m} \), in B is \( \frac{25}{m} \). So C has larger acceleration? But that seems odd. Wait, maybe the diagram is different. Wait, maybe in option C, the right force is 50 N and left is 10 N? No, the user's diagram:

A: 50 N right, 50 N left.

B: 75 N right, 50 N left.

C: 10 N right, 50 N left.

D: 100 N right, 100 N left.

Wait, maybe the mass of the cart in C is smaller? No, the problem states "these examples", implying the same cart (same mass). So according to \( a=\frac{F_{net}}{m} \), the larger the net force magnitude, the larger the acceleration. So C has net force magnitude 40 N, B has 25 N, A and D have 0. So C has the largest acceleration? But that contradicts my initial thought. Wait, maybe I made a mistake. Let's re - calculate:

Wait, net force for C: if the two forces are 10 N (right) and 50 N (left), then the net force is \( F_{net}=10 - 50=- 40\,\text{N} \) (the negative sign indicates direction, left). The magnitude is 40 N. For B, \( F_{net}=75 - 50 = 25\,\text{N} \) (right), magnitude 25 N. So 40 > 25, so C has larger net force, hence larger acceleration. But wait, maybe the question is about the direction, but acceleration magnitude is what matters. So the answer should be C? But that seems strange. Wait, maybe I misread the forces. Let me check again.

Wait, maybe in option C, the right force is 50 N and left is 10 N? No, the user's diagram:

In option C: red arrow (right) is 10 N, blue arrow (left) is 50 N.

So net force is 50 - 10 = 40 N left (magnitude 40 N).

Option B: red (right) 75 N, blue (left) 50 N. Net force 25 N right (magnitude 25 N).

So yes, C has a larger net force magnitude. So acceleration in C is larger. But wait, maybe the mass is different? The problem says "these examples", so we assume the same mass. So the answer is C? But I think I made a mistake earlier. Wait, let's check the problem again.

Wait, the question is "in which of these examples is the greatest acceleration occurring?".

Using \( a=\frac{F_{net}}{m} \), same mass, so greatest \( |F_{net}| \) gives greatest acceleration.

So:

A: \( |F_{net}| = 0 \)

B: \( |F_{net}| = 25 \)

C: \( |F_{net}| = 40 \)

D: \( |F_{net}| = 0 \)

So C has the greatest acceleration. But maybe the diagram is different. Wait, maybe I mixed up the forces in C. Maybe the right force is 50 N and left is 10 N? If that's the case, then \( F_{net}=50 - 10 = 40\,\text{N} \) right, magnitude 40 N. But the diagram shows red (right) 10 N, blue (left) 50 N. So according to the diagram, it's 10 N right and 50 N left. So net force 40 N left.

But maybe the answer is B? Wait, no, 40 is larger than 25. So unless there's a mistake in my calculation.

Wait, maybe the problem is that in option C, the cart is being pushed with a net force of 40 N, but maybe the mass is larger? No, the problem states "these examples", so same cart (same mass). So the answer should be C. But I think I made a mistake earlier. Let's confirm:

Newton's second law: \( a=\frac{F_{net}}{m} \). If \( m \) is constant, \( a\propto F_{net} \). So the largest \( F_{net} \) (magnitude) gives the largest \( a \). So C has \( F_{net}=40\,\text{N} \), B has 25 N, so C has larger acceleration.

Answer:

C