A box is given a push so that it slides across the floor. How far will it go, given that the coefficient of kinetic friction is 0.13 and the push imparts an initial speed of 3.4 m/s

s=4.44 m

Explanation:

Given that

Coefficient of the kinetic friction ,μ = 0.13

Initial velocity ,u= 3.4 m/s

Final velocity of the box ,v= 0 m/s

The acceleration due to friction force

a= - μ g

Now by putting the values in the above equation

a= - 0.13 x 10    ( take g= 10 m/s²)

 a= - 1.3 m/s²

We know that

v²= u ² + 2 a s

s=distance

a=acceleration

v=final speed

u=initial speed

Now by putting the values in the above equation

0²= 3.4² - 2 x 1.3 x s

s=4.44 m

The distance cover by box will be 4.44 m.

v=ir = 3x15 =45 v

in order to follow the instructions and complete the statements, i need to use data from table c of your student guide. can you think of a way to let me see your student guide for a while ? i can't answer the question without the data in table c.