a. Fn=F1-F2=2050-1750=300 N.,Net force.

Fn = ma.

a = Fn/m = = 300 / 1080 = 0.278 m/s^2.

b. d = Vo*t + 0.5a*t^2.

d = 0 + 0.1388*9^2 = 11.25 m.

c. V = Vo + at = 0 + 0.278*9 = 2.50 m/s.

# A boat moves through the water with two forces acting on it. One is a 2.05 103 N forward push by a motor, and the other is a 1.75 103 N resistive force due to the water.

(a) What is the acceleration of the 1080 kg boat?

(b) If it starts from rest, how far will it move in 9 s?

(c) What will its speed be at the end of its time interval?

Henry
answered

12 years ago

12 years ago

Explain Bot
answered

11 months ago

11 months ago

To solve these problems, we will use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration:

F_net = m * a

where

F_net is the net force,

m is the mass of the object, and

a is the acceleration.

(a) What is the acceleration of the 1080 kg boat?

To calculate the acceleration, we need to find the net force acting on the boat. The net force is the difference between the forward push force and the resistive force:

F_net = Forward push force - Resistive force

Substituting the given values:

F_net = 2.05 * 10^3 N - 1.75 * 10^3 N

Now we can use Newton's second law to solve for the acceleration:

F_net = m * a

a = F_net / m

Substituting the values:

a = (2.05 * 10^3 N - 1.75 * 10^3 N) / 1080 kg

Now, calculate the value of the acceleration.

(b) If it starts from rest, how far will it move in 9 s?

To determine the distance traveled, we can use the equation of motion:

d = v_i * t + 0.5 * a * t^2

where

d is the distance traveled,

v_i is the initial velocity (which is zero in this case),

t is the time, and

a is the acceleration.

Substituting the given values:

d = 0 * 9 s + 0.5 * a * (9 s)^2

Calculate the value of d.

(c) What will its speed be at the end of its time interval?

To find the final speed, we can use the equation of motion:

v_f = v_i + a * t

where

v_f is the final velocity,

v_i is the initial velocity (which is zero in this case),

a is the acceleration, and

t is the time.

Substituting the given values:

v_f = 0 + a * 9 s

Calculate the value of v_f.

F_net = m * a

where

F_net is the net force,

m is the mass of the object, and

a is the acceleration.

(a) What is the acceleration of the 1080 kg boat?

To calculate the acceleration, we need to find the net force acting on the boat. The net force is the difference between the forward push force and the resistive force:

F_net = Forward push force - Resistive force

Substituting the given values:

F_net = 2.05 * 10^3 N - 1.75 * 10^3 N

Now we can use Newton's second law to solve for the acceleration:

F_net = m * a

a = F_net / m

Substituting the values:

a = (2.05 * 10^3 N - 1.75 * 10^3 N) / 1080 kg

Now, calculate the value of the acceleration.

(b) If it starts from rest, how far will it move in 9 s?

To determine the distance traveled, we can use the equation of motion:

d = v_i * t + 0.5 * a * t^2

where

d is the distance traveled,

v_i is the initial velocity (which is zero in this case),

t is the time, and

a is the acceleration.

Substituting the given values:

d = 0 * 9 s + 0.5 * a * (9 s)^2

Calculate the value of d.

(c) What will its speed be at the end of its time interval?

To find the final speed, we can use the equation of motion:

v_f = v_i + a * t

where

v_f is the final velocity,

v_i is the initial velocity (which is zero in this case),

a is the acceleration, and

t is the time.

Substituting the given values:

v_f = 0 + a * 9 s

Calculate the value of v_f.