Answered

Get the answers you need at Westonci.ca, where our expert community is always ready to help with accurate information. Our platform provides a seamless experience for finding reliable answers from a knowledgeable network of professionals. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.


2 kg
Use 10 m/s2 for g

2 Kg Use 10 Ms2 For G class=

Sagot :

Answer:

See below

Explanation:

At point A    the PE = mgh = 2 * 10 * 1 = 20 J

  at point B, all of the PE , 20 J , is converted to Kinetic Energy

KE = 1/2 m v^2

20 = 1/2 (2)(v^2 )

20 = v^2     v = sqrt 20 = 4.47 m/s

for the friction part

 vf = vo t  + 1/2 a t^2      vf = final velocity = 0 (stopped)

                                       vo = original velocity = 4.47 m/s  

                                         a = -1 m/s^2

  0 = 4.47 t + 1/2 (-1) t^2

            - .5t^2  + 4.47 t = 0

                 t ( -.5t+ 4.47) = 0    shows t =  4.47/.5 = 8.9 seconds

Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.