. A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1.
Compute the
(a) work done by the applied force in 10 s,
(b) work done by friction in 10 s,
(c) work done by the net force on the body in 10 s,
(d) change in kinetic energy of the body in 10 s,
and interpret your results.
Answers
Answer:
Mass of the body, m = 2 kg
Applied force, F = 7 N
Coefficient of kinetic friction, µ = 0.1
Initial velocity, u = 0
Time, t = 10 s
The acceleration produced in the body by the applied force is given by Newton’s second law of motion as:
a' = F / m = 7 / 2 = 3.5 ms-2
Frictional force is given as:
f = µmg
= 0.1 × 2 × 9.8 = – 1.96 N
The acceleration produced by the frictional force:
a" = -1.96 / 2 = -0.98 ms-2
Total acceleration of the body: a' + a"
= 3.5 + (-0.98) = 2.52 ms-2
The distance travelled by the body is given by the equation of motion:
s = ut + (1/2)at2
= 0 + (1/2) × 2.52 × (10)2 = 126 m
(a) Work done by the applied force, Wa = F × s = 7 × 126 = 882 J
(b) Work done by the frictional force, Wf = F × s = –1.96 × 126 = –247 J
(c) Net force = 7 + (–1.96) = 5.04 N
Work done by the net force, Wnet= 5.04 ×126 = 635 J
(d) From the first equation of motion, final velocity can be calculated as:
v = u + at
= 0 + 2.52 × 10 = 25.2 m/s
Change in kinetic energy = (1/2) mv2 - (1/2) mu2
= (1/2) × 2(v2 - u2) = (25.2)2 - 02 = 635 J
Answer:
hey, here is your answer
mass of body, m=2 kg
applied force , f=7 n
coefficient of kinetic friction=0. 1
initial velocity=0
time=10 s
the acceleration produced in the body by applied force is given by newton's 2nd low of motion
a=f/m=7/2=3.5m/2nd
frictional force is given=0. 1*2*9. 8=-1.96
the acceleration produce by frictional force=-1. 96/2=-0. 98mper second
total acceleration of the body 3. 5+(-0. 98)=2. 52m per second
distance travelled by the body is given by equation of motion
s=ut+(1/2)at^2
=0+(1/2)*2. 52*(10) ^2=1. 26m
a) Wa=f*s=7*126=882
b) Wf=f*s=-1. 96*126=-247
c) net force=7+(-1. 96)=5. 04N
Wnet=5. 04*126=635j
d) From the equation of motion final velocity can be calculated as
v=u+at=0+2. 52*10=25. 2m/2nd
change in kinetic energy=1/2mv ^2-1/2mu^2=(1/2)*2(v^2-u^2)=(25. 2^2) -0=635