Question

Parts D and E only please

A 5 kg block is sliding at an initial speed of 9 m/s across a surface, encountering a constant friction force of 6 N. How much work is done on the block after it slides 22 cm? rrect, computer gets: -1.32 J int: Does the block gain or lose energy during this process? What sign does this imply for the work done on it? How fast is the block moving after sliding 22 cm? rrect, computer gets: 8.97 m/s int: You can treat the block like a simple particle. What kind of energy does it have and how is this affected by the work done on it? What's the total distance the block travels before coming to rest? rrect, computer gets: 33.75 What is the average power of friction on the block over the time it takes the block to come to rest? swer: Last Answer: 13.2W Submit All Answers t yet correct, tries 2/20 What was the instantaneous power of friction on the block after it slid the first 22 cm? swer: Submit All Answers t yet correct, eries 1/20

Answer 1

given
m = 5 kg
vi = 9 m/s
f = 6 N
d = 22 cm - 0.22 m

A) Workdone by friction, Wf = f*d*cos(180)

= 6*0.22*(-1)

= -1.32 J

B) Wnet = change in kinetic energy

Wnet = (1/2)*m*vf^2 - (1/2)*m*vi^2

2*Wnet/m = vf^2 - vi^2

==> vf = sqrt(vi^2 + 2*Wnet/m)

= sqrt(9^2 + 2*(-1.32)/5)

= 8.97 m/s

C) let x is the total distance tarvelled.

Wnet = KEf - KEi

f*x*cos(180) = 0 - KEi

-f*x = - (1/2)*m*v^2

x = (1/2)*m*v^2/f

= (1/2)*5*9^2/6

= 33.75 m

D) acceleration of the block during slowdown, a = -f/m

= -6/5

= -1.2 m/s^2

time taken to slowdown, t = (vf - vi)/a

= (0 - 9)/(-1.2)

= 7.5 s

Average power of friction = Workdone/time taken

= 6*33.75/7.5

= 27 W

E) instantaneous power = f*v

= 6*8.97

= 53.82 W

Note : Including sign the answers are

D) -27 W
E = -53.82 W