Calculate Applied Force

The problem involves calculating the force applied to an object based on changes in its velocity over time.

• Given Mass ($m$): $10\text{ kg}$
• Initial Velocity ($v_i$): $5\text{ m/s}$
• Final Velocity ($v_f$): $10\text{ m/s}$
• Time Duration ($\Delta t_1$): $2\text{ s}$

First, determine the change in velocity ($\Delta v$):

$ \Delta v = v_f - v_i = 10\text{ m/s} - 5\text{ m/s} = 5\text{ m/s} $

According to the Impulse-Momentum Theorem, the force applied is related to the change in momentum ($m \Delta v$) over the time interval ($\Delta t$):

$ F = \frac{m \Delta v}{\Delta t_1} $

Substitute the known values:

$ F = \frac{(10\text{ kg})(5\text{ m/s})}{2\text{ s}} = \frac{50 \text{ kg⋅m/s}}{2\text{ s}} = 25\text{ N} $

The magnitude of the applied force is $25\text{ N}$.

Calculate Final Velocity

Next, calculate the final velocity if this constant force ($F = 25\text{ N}$) acts for a different duration.

• Applied Force ($F$): $25\text{ N}$
• Mass ($m$): $10\text{ kg}$
• New Time Duration ($\Delta t_2$): $5\text{ s}$
• Initial Velocity ($v_i$): $5\text{ m/s}$ (Assuming the scenario starts from the same initial condition)

Calculate the acceleration ($a$) caused by the force:

$ a = \frac{F}{m} = \frac{25\text{ N}}{10\text{ kg}} = 2.5\text{ m/s}^2 $

Use the kinematic equation to find the final velocity ($v_{f2}$):

$ v_{f2} = v_i + a \Delta t_2 $

Substitute the values:

$ v_{f2} = 5\text{ m/s} + (2.5\text{ m/s}^2)(5\text{ s}) $

$ v_{f2} = 5\text{ m/s} + 12.5\text{ m/s} = 17.5\text{ m/s} $

The final velocity after applying the force for $5\text{ s}$ is $17.5\text{ m/s}$.

Final Result

The applied force is 25 N, and the final velocity after 5 seconds is 17.5 m/s.