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?

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

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

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

= 0 + 2(4.71 x 12)

 = 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

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

= + 2(3.72 x 10.5)

 = 112.9 + 78.12

v  =  = 13.82 m/s

the answer would be 32*f