Question

Please answer the four questions below. I attached the previous post.

A 200. kg iceboat is accelerating in the +x direction on a frictionless frozen lake.

->a) What is the iceboat's acceleration in m/s²?

->b) At t = 0, the ice boat has an initial velocity of 2.00 m/s. What was the iceboat's initial momentum? (in kg m/s)

->c) What was the iceboat's initial kinetic energy? (in Joules)

->d) What is the kinetic energy after going through the 20.0 m displacement? (in Joules)

A) 1.00 x 10², B) 2.00 x 10², C) 4.00 x 10², D) 1.04 x 10³, or E) 1.04 x 10⁴

^{-------------------------------------------------------------}

Previously post:

A 200. kg iceboat is accelerating in the +x direction on a frictionless frozen lake.

1. There is a net force acting on the iceboat. True or False

2. There could be one or more forces, parallel to the icy surface, acting on the iceboat. True or False

3. If the net force in the x-direction is +500. N, how much work does the net force do over a displacement of 20.0 m? (in units of Joules)

A) 200., B) 500., C) 1.00 x 10³, D) 1.00 x 10⁴, E) none of these

4)The kinetic energy doesn't change over the 20.0 m displacement. True or False

5) What is the iceboat's acceleration in m/s²?

6) At t = 0, the ice boat has an initial velocity of 2.00 m/s. What was the iceboat's initial momentum? (in kg m/s)

7) What was the iceboat's initial kinetic energy? (in Joules)

8) What is the kinetic energy after going through the 20.0 m displacement? (in Joules)

A) 1.00 x 10², B) 2.00 x 10², C) 4.00 x 10², D) 1.04 x 10³, or E) 1.04 x 10⁴

⁹⁾ Over the 20.0 m displacement, the net non-conservative work was zero. True or False

10) If there had been a frictional force acting on the iceboat in addition to the +500. N force, the frictional force would have opposed the direction of motion. True or False

11) The normal force on the iceboat is 200. kg x 9.80 m/s² = 1.96 x 10³ N. True or False

Answer 1

Mass of the iceboat, m = 200 kg

a. Acceleration, a = force/mass = 500/200 = 2.5 m/s²

b. Initial velocity, u = 2 m/s. initial momentum, p = mu = 200 x 2 = 400 kgm/s

c. Initial kinetic energy, E1 = ½ mu² = ½ x 200x 2 x 2 = 400 J

d. Change in kinetic energy = work done

E2 – E1 = W

Final kinetic energy, E2 = W + E1 = 10000 + 400 = 10400 J = 1.04 x 10⁴ J

( E ) 1.04 x 10⁴ J