Question

The figure below shows a cold package of hot dogs sliding rightward across a frictionless floor through a distance d = 18.0 cm while three forces are applied to it.Two of the forces are horizontal and have the magnitudes F1 = 5.00 N and F2 = 1.00 N; the third force is angled down by ? = 70.0° and has the magnitude F3 = 4.00N.
(a) For the 18.0 cm displacement, what is the net work done on the package by the three applied forces, the gravitational force on the package, and the normal force onthe package?
http://www.webassign.net/hrw/W0132-N.jpg
J

(b) If the package has a mass of 1.9 kg and an initial kinetic energy of 0, what is its speed at the end of the displacement?
m/s

Answer 1

Answer 2

Net horizontal force

  F_x   = (F₁  + F_3x - F₂)

where F_3x = F₃ * cos θ

F_x   = 5.00 + 4.00 * cos 70⁰ - 1.00

= 2.63 N

gravitational force F_g  = m * g

normal force F_y = F_g  + F_3y

= m * g  + F₃ * sin θ = m * g + 4.00 * sin 70⁰

= mg + 3.76 N

a. Work done by applied forces W_a   = F_x * d

(since only component of force that are in direction of displacement do work, I have taken only x component)

W_a   = 2.63 * 18 cm

= 2.63 * 0.18

= 0.4734 J

Work done by gravitational force W_g   = m * g * d * cos α

= m * g * d * cos 90⁰

= 0

Work done by normal forces W_n   = F_y * d * cos β

= (m * g + 3.76) * cos 90⁰

= 0

b. Net work done W = W_a + W_g + W_n

= 0.4734 + 0 + 0

= 0.4734 J

Final k.e. K_f   = K_i   + W ( work energy theorem)

(1/2) * m * v²   = 0 + 0.4734

v = √(2 * 0.4734 / 1.9)

= 0.7059 m/s