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

Question

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.

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Thạch Thảo 5 years 2021-07-23T06:18:42+00:00 1 Answers 306 views 1

Answers ( )

  1. Sigrid
    0
    2021-07-23T06:20:29+00:00
    Reply

    Answer:

    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

               

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