a 3.80 kg fish swims upstream against the current. in order to make it over a rock jutting out of the water…

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?

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?

Answer

Explanation:

Step1: Find the weight of the fish

The mass of the fish ( m = 3.80\space kg ), and the acceleration due to gravity ( g = 9.8\space m/s^2 ). The weight ( W = mg ), so ( W = 3.80\times9.8 = 37.24\space N ).

Step2: Determine the net force

We have horizontal force ( F_h = 131\space N ) and vertical force ( F_v = 97.0\space N ), and weight (downward force) ( W = 37.24\space N ). The net vertical force ( F_{net,v}=F_v - W=97.0 - 37.24 = 59.76\space N ). The net horizontal force ( F_{net,h}=131\space N ). The magnitude of the net force ( F_{net}=\sqrt{F_{net,h}^2 + F_{net,v}^2}=\sqrt{131^2 + 59.76^2}=\sqrt{17161+3571.2576}=\sqrt{20732.2576}\approx143.99\space N )

Step3: Calculate the acceleration

Using Newton's second law ( F = ma ), so ( a=\frac{F_{net}}{m} ). Substituting ( F_{net}\approx143.99\space N ) and ( m = 3.80\space kg ), we get ( a=\frac{143.99}{3.80}\approx37.9\space m/s^2 )

Answer:

The acceleration of the fish just before it jumps is approximately ( \boldsymbol{38\space m/s^2} ) (or more precisely ( 37.9\space m/s^2 ))