Question

An object of mass 3.60 kg undergoes a displacement S= 12.2 m i + 7.9 m j while it is being acted on by a net force F = 9.0 N i -1.5 N j. How much work is done on the object by the net force? Also If the object above starts from rest, what is its speed after it has undergone the displacement described above?

Answer 1

use W = F.S

Answer 2

dont forget to use dot vector identity

Answer 3

W = F.s
W = (9.0 Ni - 1.5 Nj) . (12.2 mi + 7.9 mj)
W = ((9.0 x 12.2)i.i - (1.5x7.9)j.j
W = (109.8 - 11.85) N.m

Answer 4

thank you but how would i put this in loncapa

Answer 5

Then, F = m.a

F = m.a
a = F/m
\[a =\frac{(9.0 Ni - 1.5 Nj) }{3.60 }\]
a = 2.5 N/m i - 0.42 N/m j

subtitutes a to vt^2 = vo^2 + 2 a. s,
where vt is the final speed, vo is the initial speed, a is acceleration,

vt^2 = vo^2 + 2 a. s,
vt^2 = 0 + 2 a. s,
vt^2 = (2.5 N/m i - 0.42 N/m j) . (12.2 m i + 7.9 m j )

Answer 6

but i'm sorry I mistakenly wrote:

for a unit not N/m but N/kg

Answer 7

Couldn't you also use kinetic energy to find the speed?
That seems more direct than using F=ma and kinematics.

Answer 8

vt^2 = 2 .(2.5 N/kg i - 0.42 N/kg j) . (12.2 m i + 7.9 m j )
vt^2 = 2 .(30.5 - 3.318)
vt = sqrt (2.(30.5 - 3.318)) m/s

Answer 9

yes i could...,

Answer 10

W = delta EK

Answer 11

W = 1/2. mv^2

Answer 12

Makes the algebra a little easier, eh?

Answer 13

sure