if you are copying answers from somewhere else plz dont answer it im losing poin

Question

if you are copying answers from somewhere else plz dont answer it im losing points and the dead line is soon A popular pastime is to see who can push an object closest to the edge of a table without its going off. You push the 100 g object and release it 1.00m from the table edge. Unfortunately, you push a little too hard. The object slides across, sails off the edge, falls 0.700m to the floor, and lands 40.0cm cm from the edge of the table. If the coefficient of kinetic friction is 0.500, what was the object's speed as you released it?

Explanation / Answer

Consider the equation

s=ut+1/2at^2

by this we compute time

0.7=0.5*9.8*t^2

SO

t=0.37 seconds

now calculate velocity

0.37v=0.40

v=1.081 m/s

Frictional Force= 0.5*0.1*9.8=0.49N

Now we know that

F=ma

0.49=0.1a

a=4.9 m/s^2

so using equation

v^2=u^2+2as

1.081^2=u^2-2*4.9*1

u=3.31 m/s

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