Slowing down boats-have answer need help getting to it!

Question

Sam’s job at the amusement park is to slow down and bring to a stop the boats in the log ride. The boat and its riders have a mass of 910kg . If he applies a constant 445N force and stops the boat in 2.0 m, how fast was the boat initially going?

Answer: 1.4 m/s

Answer 1

This could be solved using the work-energy theorem.

W_net = ΔKE = KE_f - KE_i

Since it comes to a stop, v_f = 0 and so KE_f = 0

W_net = -KE_i = -(1/2)mv²

W_net = Fdcosθ

θ = angle between F vector and d vector. Since the object is slowing down, the force must be applied opposite to the displacement, so θ = 180 deg and cosθ = -1

W_net = -Fd = -(445 N)(2.0 m) = -890 J

v = sqrt(-2W_net/m) = sqrt (-2(-890 J)/910 kg)) = 1.4 m/s

Answer 2

To start, consider what information you are given and what unknowns you are being asked to find.

GIVEN:

mass (m) = 910kg

force (F) = -445N (negative because it is applied in the opposite direction of the velocity

displacement (d) = 2.0m

final velocity (Vf) = 0 m/s (because the boat comes to a complete stop at the end)

UNKNOWN:

initial velocity (Vo) = ?

acceleration (a) = ?

Finding the acceleration is simple, as we already have the mass of the boat and the force applied to stop it.

F = ma

(-445N) = (910kg)a

a = -445N/910kg

a = -0.49m/s^2

Now we need to find the value of the initial velocity. Take a moment and look through your acceleration equations, which one uses all of the values we now know from this problem?

Vf^2 = Vo^2 + 2ad

(0m/s)^2 = Vo^2 + 2(-0.49m/s^2)(2m)

Vo = √((0m/s^2) - (2(-0.49m/s^2)(2m))

Vo = 1.4m/s