A 0.50 kg skateboard is at rest on a rough, level floor on which two lines have been drawn 1.0 m
apart. A constant horizontal force is applied to the skateboard at the beginning of the interval,
and is removed at the end. The skateboard takes 8.5 s to travel the 1.0 m distance, and it then
coasts for another 1.25 m before coming to rest. Calculate the force applied to the skateboard,
and also the constant frictional force opposing its motion.

Ff = 0.01107 N - Frictional force

P = 0.02491 N - Applied force

Explanation:

Given:-

- The mass of the skateboard, m = 0.50 kg

- The first part of journey was for distance, s1 = 1.0 m

- The time duration of first part of journey, t1 = 8.5 s

- The second part of the journey was for distance, s2 = 1.25 m

- The initial and final conditions of the skateboard are at rest.

Find:-

Calculate the force applied to the skateboard,  and constant frictional force opposing its motion.

Solution:-

- The first part of the journey had two forces acting on the skateboard over the distance "s1" and time duration "t1".

- The applied force " P " makes the skateboard move and the opposing frictional force " Ff " opposes the motion.

- Assuming the linear acceleration " a " of the skateboard to remain constant.

- Apply the second kinematic equation of motion:

                        s1 = u*t1 + 0.5*a*t1^2

Where,

          u: The initial velocity.  ( skateboard is at rest; hence, 0 )

                       s1 = 0.5*a*t1^2

                       a = 2*s1 / t1^2

                       a = 2*1 / 8.5^2

                       a = 0.02768 m/s^2

- Similarly, the second part of the journey devoid of the applied force " P " and only the frictional force "Ff" acts on the block slowing the skateboard to rest.

- Apply the first equation of motion to determine the speed "v1" as soon as the force "P" is removed or the start of the second journey.

                      v1 = u + a*t

                      v1 = 0 + 0.02768*8.5

                      v1 = 0.23528 m/s

- The block slows down from the speed of v = 0.23528 to zero over the distance of s2 = 1.25. The linear acceleration for the second part of the journey ( a2 ) can be determined from the 3rd equation of motion:

                     v2^2 = v1^2 + 2*a2*s2

Where,

          v2: The final velocity of second journey ( rest ) = 0

                    0 = 0.23528^2 + 2*a2*1.25

                    a2 = - 0.23528^2 / 2*1.25

                    a2 = - 0.02214 m/s^2

- The second part of the journey the skateboard decelerates to rest with linear acceleration of "a2".

- Apply the Newton's second law of motion for the second part of the journey:

                   F_net = m*a2

Where,

          F_net = - Ff ( Frictional force )

                  -Ff = 0.5*( -0.02214 )

                   Ff = 0.01107 N  ... Answer

- Apply the Newton's second law of motion for the first part of the journey:

                  F_net = m*a

Where,

          F_net = P - Ff ( Both forces act )

                  P - Ff = 0.5*( 0.02768 )

                  P = 0.01384 + 0.01107

                  P = 0.02491 N   Answer