Question

6. A force of 40 N is applied to a 6 kg box of books. If the frictional force acting on the box is 16

N, and the box was initially at rest, what is the velocity of the box 10 s later? What is the box's
displacement over the 10 s period?

Answer 1

V = 40 m/s^2

Step-by-step explanation:

horizontal force ,
`F_h = 40N`

mass, m = 6kg

time, t = 10s

frictional force,
`F_f = 16N`

net force
`F = F_h - F_f = ma`

40N - 16N = 6 x a (1)

in this case you apply the equation of motion

the block is at rest, initial velocity is, U = 0

final velocity is is what we are looking for, V

`a = (V - U)/(t)`

since U is 0 we have

a = V/t (2)

subsituting (2) into (1)

`24 = (6*V)/(t)`

`24 = (6*V)/(10)`

cross multiply

240 = 6 x V

divide both sides by 6

V = 40 m/s^2