Question

two people are paddling a row boat. together they exert a horizontal force of 630 n toward the back of the boat. the water resists the motion of the boat with a force of 200 n. if the combined mass of the boat and the two people in it is 215 kg, what is the acceleration of the rowboat?
1.47 m/s^2, forward
2.93 m/s^2, forward
2.01 m/s^2, forward
0.683 m/s^2, forward

in this image f1 is equal to 150 n and f2 is equal to 100 n. what is the net force?
image of two people pushing a table
fnet = 50 n to the right
fnet = 0 n
fnet = 50 n to the left

Explanation:

First Sub - Question (Rowboat Acceleration)

Step1: Determine Net Force

The force exerted by the people is \( F_{people}=630\ N \) (backward), and the resistive force of water is \( F_{water}=200\ N \) (opposing the motion, so if we take forward as positive, the net force \( F_{net}=F_{people}-F_{water} \)? Wait, no. Wait, when they paddle backward, the boat moves forward (Newton's third law). So the force that propels the boat forward is equal to the force they exert backward (action - reaction). Wait, no, let's clarify the directions. Let's assume the forward direction is positive. The people exert a force of 630 N backward, so the boat experiences a force of 630 N forward (because when you push water backward, water pushes you forward). The water resists with a force of 200 N backward (opposing the motion). So the net force \( F_{net}=630\ N - 200\ N=430\ N \) (forward).

Step2: Apply Newton's Second Law

Newton's second law is \( F = ma \), where \( F \) is the net force, \( m \) is the mass, and \( a \) is the acceleration. We know \( m = 215\ kg \) and \( F_{net}=430\ N \). So we can solve for \( a \): \( a=\frac{F_{net}}{m}=\frac{430\ N}{215\ kg}=2.0\ m/s^{2}\approx2.01\ m/s^{2} \) (forward).

Second Sub - Question (Net Force on the Table)

From the diagram, we can see the direction of the forces. Let's assume the right - hand side force (from the person in red) is \( F_1 = 150\ N \) to the right and the left - hand side force (from the person in white) is \( F_2 = 100\ N \) to the left. The net force \( F_{net}=F_1 - F_2 \) (since they are in opposite directions). So \( F_{net}=150\ N-100\ N = 50\ N \), and the direction is to the right (because \( F_1>F_2 \)).

Answer:

2.01 m/s², forward