Question

A 51.0 kg box, starting from rest, is pulled across a floor with a constant horizontal force of 240 N. For the first 12.0 m the floor is frictionless, and for the next 10.5 m the coefficient of friction is 0.21. What is the final speed of the crate after being pulled these 20.5 meters?

Answer 1

The final speed of the crate after being pulled these 20.5 meters is 13.82 m/s

Step-by-step explanation:

I'll assume that the correct question is

A 51.0 kg box, starting from rest, is pulled across a floor with a constant horizontal force of 240 N. For the first 12.0 m the floor is frictionless, and for the next 10.5 m the coefficient of friction is 0.21. What is the final speed of the crate after being pulled these 22.5 meters?

mass of box = 51 kg

for the first 12 m, it is pulled with a constant force of 240 N

The acceleration of the box for this first 12 m will be

from F = ma

a = F/m

where F is the pulling force

m is the mass of the box

a is the acceleration of the box

a = 240/51 = 4.71 m/s^2

Since the body started from rest, the initial velocity u = 0

applying Newton's equation of motion to find the final velocity at the end of the first 12 m, we have

`v^(2)= u^(2)+2as`

where v is the final velocity

u is the initial velocity which is zero

a is the acceleration of 4.71 m/s^2

s is the distance covered which is 12 m

substituting value, we have

`v^(2)` = 0 + 2(4.71 x 12)

`v^(2)` = 113.04

`v = √(113.04)` = 10.63 m/s

For the final 10.5 m, coefficient of friction is 0.21

from f = μF

where f is the frictional force,

μ is the coefficient of friction = 0.21

and F is the pulling force of the box 240 N

f = 0.21 x 240 = 50.4 N

Net force on the box = 240 - 50.4 = 189.6 N

acceleration = F/m = 189.6/51 = 3.72 m/s^2

Applying newton's equation of motion

`v^(2)= u^(2)+2as`

u is initial velocity, which in this case = 10.63 m/s

a = 3.72 m/s^2

s = 10.5 m

v = ?

substituting values, we have

`v^(2)` =
`10.63^(2)` + 2(3.72 x 10.5)

`v^(2)` = 112.9 + 78.12

v =
`√(191.02)` = 13.82 m/s