Question

A mysterious rocket-propelled object of mass 49.0 kg is initially at rest in the middle of the horizontal, frictionless surface of an ice-covered lake. Then a force directed east and with magnitude F(t)=(16.3 N/s)t is applied. How far does the object travel in the first 5.50s after the force is applied?

Answer 1

The object travels approximately 11.548 m in the first 5.50s after the force is applied.

Step-by-step explanation:

To find the distance traveled by the object in the first 5.50s after the force is applied, we need to calculate the acceleration using Newton's second law and then use the kinematic equation for distance. The force applied is given by F(t) = (16.3 N/s)t. Since the force is changing with time, we need to find the average force over the given time period.

First, we find the average force using the equation:

F_avg = (1/t) ∫[t=0 to t=5.50] F(t) dt

Integrating the equation, we get:

F_avg = (1/t) ∫[t=0 to t=5.50] (16.3 N/s)t dt

F_avg = (1/5.50) ∫[t=0 to t=5.50] (16.3 N/s)t dt

F_avg = (1/5.50) [(16.3 N/s) * (5.50^2/2) - (16.3 N/s) * (0^2/2)]

F_avg = (1/5.50) [(16.3 N/s) * (5.50^2/2)]

F_avg = (1/5.50) * (16.3 N/s) * (5.50^2/2)

F_avg = 37.407 N

Now that we have the average force, we can use Newton's second law to find the acceleration:

F_avg = m * a

a = F_avg / m

a = 37.407 N / 49.0 kg

a = 0.763 m/s^2

Now that we have the acceleration, we can use the kinematic equation for distance:

d = d₀ + v₀t + (1/2)at²

Since the object is initially at rest, d₀ = 0

Substituting the values:

d = (1/2)(0.763 m/s²)(5.50 s)²

d = (1/2)(0.763 m/s²)(30.25 s²)

d = (0.3815 m/s²)(30.25 s²)

d ≈ 11.548 m

Therefore, the object travels approximately 11.548 m in the first 5.50s after the force is applied.