Momentum = m • v
Original momentum = m • 10 m/s north
Final momentum = m • 15 m/s north
Change = m • (15 - 10) m/s north
Change = m • +5 m/s north
Change = +60 kg-m/s north
P=w/t
so the answer is 500/10=50
Let:
Vx = the pulling component of force
Vy = the lifting component of force
Vy:
Sin(n°) = Vy/hypotenuse
hypotenuse * Sin(n°) = Vy
100N*sin(30°) = Vy
50N = Vy
Vx:
Cos(n°) = Vx/hypotenuse
Hypotenuse * cos(n°) = Vx
100N*cos(30°) =Vx
about 86.6N = Vx
Answer:
v=39.05 m/s
Explanation:
Given that
x= 56 cm
F= 158 N
m= 58 g = 0.058 kg
Lets take spring constant = k
At the initial position,before releasing the arrow
F= k x
By putting the values
F= k x
158= 0.56 k
k=282.14 N/m
Now from energy conservation
Lets take final speed of the arrow after releasing

k x²=mv²
282.14 x 0.56² = 0.058 v²
v=39.05 m/s