A 30.0kg package is moving along a conveyer bel a 0.500m/s to the right when an

Question

A 30.0kg package is moving along a conveyer bel a 0.500m/s to the right when an applied force 120N is used to push the package to the right. At the same time friction acs on the package wih 5.00N as also shown in the diagam. When the package has move a distance of 0.800 m,, what is the speed of the package? (follow the steps)a) what is the initial kinetic energy. b) What is the net work done on the package? c) What is the final kinetic energy of the package? d) what is the final speed of the package?

Explanation / Answer

1. Net force acting on package = Force applied - friction = 120 - 5 = 115 N
Through distance 0.8 m

a) Initial KE = 0.5*30 *0.5*0.5 = 3.75 J

b) Work done = Fx = 115 * 0.8 = 92 J

c) Final KE = 92+3.75 = 95.75 J

d) From final KE, V*V where V is final velocity is 95.75*2/30
or V = 2.53 m/s

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