A block moving with initial speed of 12.0 m/ enters a frictionlessplane making a

Question

A block moving with initial speed of 12.0 m/ enters a frictionlessplane making a 60°-angle with the vertical. a) find the distance traveled by the block along the planebefore its seed turns into zero
b) The same enters a rough horizontal surface having acoefficient of kinetic friction of 0.21. What is the speed of theblock after it travels 5.0 m along the surface? a) find the distance traveled by the block along the planebefore its seed turns into zero
b) The same enters a rough horizontal surface having acoefficient of kinetic friction of 0.21. What is the speed of theblock after it travels 5.0 m along the surface?

Explanation / Answer

Given that The initial speed of the block (u) =12.0m/s The angle made with the vertical () =600 The final speed (v) =0 m/s Acceleration due to gravity (g) =9.8m/s2 a) As we know that the block is moving in the upward directionthen the acceleration is given by                              a =gsin Now from the equation of motion weget that                          v2 -u2 =2as The distance traveled by the block along the plane before itsseed turns into zero is                               s = v2-u2/2gsin                                  = 8.48m b) The initial speed of the block (u) =120.0m/s The coefficient of kinetic friction of(k) =0.21 The block moved to a distance(s) = 5.0m Acceleration due to gravity (g) =9.8m/s2 Now for the rough horizontal surface the acceleration is givenby                            a = -kg Now the final speed of the block after traveling through adistance on the rough surface is given by                            v2-u2 =2as Now we get that                            v2 = u2 + 2as Then the final speed (v) = u2+2as                                         = 144 -20.58                                    =11.11m/s               The final speed (v) =0 m/s Acceleration due to gravity (g) =9.8m/s2 a) As we know that the block is moving in the upward directionthen the acceleration is given by                              a =gsin Now from the equation of motion weget that                          v2 -u2 =2as The distance traveled by the block along the plane before itsseed turns into zero is                               s = v2-u2/2gsin                                  = 8.48m b) The initial speed of the block (u) =120.0m/s The coefficient of kinetic friction of(k) =0.21 The block moved to a distance(s) = 5.0m Acceleration due to gravity (g) =9.8m/s2 Now for the rough horizontal surface the acceleration is givenby                            a = -kg Now the final speed of the block after traveling through adistance on the rough surface is given by                            v2-u2 =2as Now we get that                            v2 = u2 + 2as Then the final speed (v) = u2+2as                                         = 144 -20.58                                    =11.11m/s              

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