A duck has a mass of 2.00 kg. As the duck paddles, a force of 0.130 N acts on it

Question

A duck has a mass of 2.00 kg. As the duck paddles, a force of 0.130 N acts on it in a direction due east. In addition, the current of the water exerts a force of 0.220 N in a direction of 56.0° south of east. When these forces begin to act, the velocity of the duck is 0.110 m/s in a direction due east. Find (a) the magnitude and (b) the direction (relative to due east) of the displacement that the duck undergoes in 2.00 s while the forces are acting. (Note that the angle will be negative in the south of east direction.)

Explanation / Answer

I will be using î as the unit vector in the East direction and as the unit vector in the North direction; West and South are the negatives of these, respectively.

Let m = the mass of the duck
Let A = the acceleration vector of the duck
Let F1 = the first force vector acting on the duck = (0.130 N)î
Let F2 = the second force vector acting on the duck = (0.220 N){cos(-56°)î + sin(-56°)}
Let V0 = the initial velocity vector = (0.110 m/s)î
Let V(t) = the velocity vector
Let S(t) = the displacement vector (assuming that S(0) = 0)

mA = F1 + F2

A = F1/m + F2/m

A = [(0.13 N)î + (0.220 N){cos(-56°)î + sin(-56°)}]/(2 kg)

Grouping î and the terms together:

A = {(0.13 N) + (0.220 N)cos(-56°)}î/(2 kg) + (0.220 N)sin(-56°)/(2 kg)

Simplifying:

A = (0.127 m/s²)î - (0.091 m/s²)

V(t) = At + V0

V(t) = (0.127 m/s²)tî - (0.091 m/s²)t + (0.110 m/s)î

Grouping like terms together:

V(t) = {(0.127 m/s²)t + (0.110 m/s²)}î - (0.091 m/s²)t

S(t) = (1/2)At² + V0t

S(t) = {(0.127 m/s²)t²/2 + (0.110 m/s)t}î - (0.091 m/s²)t²/2

S(2 s) = {(0.127 m/s²)(2 s)²/2 + (0.110 m/s)(2 s)}î - (0.091 m/s²)(2 s)²/2

S(2.20 s) = (0.474 m)î - (0.182 m)

The magnitude is:

|S(2.20 s)| = {(0.474 m)² + (0.182 m)²} .2247+

|S(2.20 s)| = 0.51 m

The direction is:

= tan^-1(-0.182/0.474)

= -21°

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