Question

a 3.80 kg fish swims upstream against the current. in order to make it over a rock jutting out of the water, the fish needs to accelerate upwards at an angle and then jump the rock barrier. the fish needs to both overcome the horizontal force of the current, which is 131 n, as well as make it over the rocks with a vertical force of 97.0 n. what is the acceleration of the fish just before it makes its jump?

Explanation:

Step1: Find the net force magnitude

First, we know the horizontal force \( F_x = 131\space N \) and vertical force \( F_y = 97.0\space N \). The net force \( F_{net} \) can be found using the Pythagorean theorem: \( F_{net}=\sqrt{F_x^{2}+F_y^{2}} \)

$$ F_{net}=\sqrt{(131)^{2}+(97.0)^{2}}=\sqrt{17161 + 9409}=\sqrt{26570}\approx163\space N $$

Step2: Use Newton's second law to find acceleration

Newton's second law is \( F = ma \), so \( a=\frac{F_{net}}{m} \). The mass \( m = 3.80\space kg \)

$$ a=\frac{163}{3.80}\approx42.9\space m/s^{2} $$

(We can also calculate the direction, but the question asks for acceleration, and if we consider magnitude and direction, but let's check the force components again. Wait, actually, the forces here are the net forces? Wait, no, the horizontal force is the force to overcome current, vertical force to overcome... Wait, maybe the horizontal force is the applied force horizontally, vertical force applied vertically. So the net force is the vector sum of these two forces. Then mass is 3.80 kg. So \( F_{net}=\sqrt{(131)^2+(97)^2} \approx \sqrt{17161 + 9409}=\sqrt{26570}\approx163\space N \). Then \( a = F_{net}/m = 163/3.80\approx42.9\space m/s^{2} \). Wait, but let's check the calculation again. \( 131^2 = 17161 \), \( 97^2 = 9409 \), sum is 26570, square root is approx 163. So 163 / 3.8 ≈ 42.9 m/s². Alternatively, maybe we should consider that the horizontal force is the force the fish applies horizontally, vertical force vertically, so net force is the vector, then acceleration is net force over mass.

Wait, another way: horizontal acceleration \( a_x = F_x/m = 131/3.80 ≈ 34.47\space m/s² \), vertical acceleration \( a_y = F_y/m = 97.0/3.80 ≈ 25.53\space m/s² \). Then the magnitude of acceleration is \( \sqrt{a_x^{2}+a_y^{2}} \)

$$ a=\sqrt{(34.47)^{2}+(25.53)^{2}}=\sqrt{1188 + 651}=\sqrt{1839}\approx42.9\space m/s^{2} $$

Same result. So the acceleration magnitude is approximately \( 42.9\space m/s^{2} \) (and direction can be found by \( \theta=\arctan(\frac{a_y}{a_x})=\arctan(\frac{97}{131})\approx36.5^\circ \) above horizontal, but the question says "acceleration", maybe magnitude and direction, but let's see the problem statement. It says "acceleration of the fish just before it makes its jump", so we can give the magnitude or the vector. But likely magnitude. Wait, let's recalculate \( F_{net} \):

\( 131^2 = 17161 \), \( 97^2 = 9409 \), sum is 26570, square root is 163.0 (since 163^2 = 26569, which is very close to 26570, so 163 N). Then \( a = 163 / 3.80 ≈ 42.9\space m/s² \). So that's the acceleration magnitude.

Answer:

The acceleration of the fish is approximately \(\boxed{42.9\space m/s^{2}}\) (magnitude; the direction is an angle \(\theta=\arctan(\frac{97.0}{131})\approx36.5^\circ\) above the horizontal, but if only magnitude is needed, this is it).