Answer:
A) 667 J
B) 381.4 J
C) 0 J
D) 245.4 J
E) 40.2J
F) 2 m/s
Explanation:
Let g = 9.81 m/s2
A) The work done on the suitcase is the product of the force applied and the distance travelled:
w = Fs = 145 * 4.6 = 667 J
B) The work done by gravitational force the dot product between the gravity vector and the distance vector
[tex]W_g = \vec{P}\vec{s} = mgs sin\alpha = 20*9.81*4.6*sin25^o = 381.4 J[/tex]
C) As the normal force vector is perpendicular to the distance vector, the work done by the normal force is 0
D) The work done on the suitcase by friction force is the product of the force applied and the distance travelled, whereas friction force is the product of normal force and coefficient
[tex]W_f = F_fs = \mu N s = \mu s mgcos\alpha = 0.3* 4.6 * 20*9.81*cos25^o = 245.4 J[/tex]
E) The total workdone on the suite case would be the pulling work subtracted by gravity work and friction work
[tex]W = w – W_g – W_f = 667 – 381.4 – 245.4 = 40.2 J[/tex]
F) As the suit case has 0 kinetic and potential energy at the bottom, and the total work done is converted to kinetic energy at 4.6 m along the ramp, we can conclude that:
[tex]E_k = W = 40.2 j[/tex]
[tex]mv^2/2 = 40.2 [/tex]
[tex]20v^2/2 = 40.2[/tex]
[tex]10v^2 = 40.2[/tex]
[tex]v^2 = 4.02[/tex]
[tex]v = \sqrt{4.02} = 2 m/s[/tex]
