Question

In a pickup game of dorm shuffleboard, students crazed by final exams use a broom to propel a calculus book along the dorm hallway. If the 3.5 kg book is pushed from rest through a distance of 0.78 m by the horizontal 26 N force from the broom and then has a speed of 1.68 m/s, what is the coefficient of kinetic friction between the book and floor?

Answer 1

Answer:
`\mu =0.573`

Step-by-step explanation:

Given

mass of book
`m =3.5 kg`

distance moved by book
`s=0.78 m`

Force applied
`F=26 N`

initial speed
`u=1.68 m/s`

coefficient of kinetic friction is
`\mu`

using
`v^2-u^2=2 as`

`a=(0-1.68^2)/(2* 0.78)`

`a=1.809 m/s^2` deceleration

Net Force

`F-f_r=ma`

`F-ma=f_r` , where
`f_r=friction\ force`

`f_r=\mu N=\mu mg`

`26-3.5* 1.809=\mu \cdot 3.5* 9.8`

`\mu =(26-6.33)/(34.3)`

`\mu =(19.67)/(34.3)=0.573`

Answer 2

μ=0.16

Step-by-step explanation:

Given that

m= 3.5 Kg

d= 0.78 m

F= 26 N

v= 1.68 m/s

Lets take coefficient of kinetic friction = μ

Friction force Fr=μ m g

Lets take acceleration of block is a m/s²

F- Fr = m a

26 - μ x 3.5 x 10 = 3.5 a ( take g =10 m/s²)

a= 7.42 - 35μ m/s²

The final speed of the block is v

v= 1.68 m/s

We know that

v²= u²+ 2 a d

u= 0 m/s given that

1.68² = 2 x a x 0.78

a= 1.80 m/s²

a= 7.42 - 35μ m/s²

7.42 - 35μ = 1.80

μ=0.16