Question

A box of mass 18.0 kg sits at rest on a horizontal surface. The coefficient of kinetic friction between the surface and the box is 0.300. The box is initially at rest, and then a constant force of magnitude FF and direction 39.0 ∘∘ below the horizontal is applied to the box; the box slides along the surface.

A. What is F if the box has a speed of 6.00 m/s after traveling a distance of 8.00 mm? Express your answer with the appropriate units.

B. What is F_B if the surface is frictionless and all the other quantities are the same? Express your answer with the appropriate units.

C. What is F_C if all the quantities are the same as in part A but the force applied to the box is horizontal? Express your answer with the appropriate units.

Answer 1

let
m = 18 kg
theta = 39 degrees
d = 8 m
vf = 6 m/s

A) Normal force acting on the block,

N = m*g + F*sin(39)

Net force acting along horizontal direction,

Fnet = F*cos(39) - fk

= F*cos(39) - N*mue_k

= F*cos(39) - (m*g + F*sin(39))*mue_k

now use, Work-energy theorem,

Net workdone = change in kinetic energy

Fnet*d = KEf - KEi

(F*cos(39) - (m*g + F*sin(39))*mue_k )*d = (1/2)*m*v^2 - 0

F*cos(39) - m*g*mue_k - F*sin(39)*mue_k = m*v^2/(2*d)

F*(cos(39) - sin(39)*mue_k) = m*v^2/(2*d) + m*g*mue_k

F = (m*v^2/(2*d) + m*g*mue_k )/(cos(39) - sin(39)*mue_k)

= (18*6^2/(2*8) + 18*9.8*0.3)/(cos(39) - sin(39)*0.3)

= 159 N <<<<<<<---------------Answer

B) Net workdone = gain in kinetic energy

FB*d*cos(39) = (1/2)*m*v^2 - 0

FB = m*v^2/(2*d*cos(39))

= 18*6^2/(2*8*cos(39))

= 52.1 N <<<<<<<---------------Answer

C)

FC = (m*v^2/(2*d) + m*g*mue_k )/(cos(0) - sin(0)*mue_k)

= (18*6^2/(2*8) + 18*9.8*0.3)/(cos(0) - sin(0)*0.3)

= 93.4 N <<<<<<<---------------Answer