A body of mass 2 kg initially at rest moves under the action of applied horizontal force of 7 newton on a table with coefficient of kinetic friction 0.1 calculate work done by force in 10 seconds work done by friction in 10 second work done by net force on body in 10 second change in kinetic energy of body in 10 sec
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.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
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
prachisharma4:
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mass of body = 2 kg
applied force= 7 N
coefficient of kinetic friction= 0.1
initial velocity/u= 0
time,t= 10sec.
The acceleration produced in the body by applied force is given by Newton's second law of motion as : acceleration(a)= f/m =7/2=
Now,
friction force given as: 0.1 ×2×9.8 =1.96N
the acceleration produced by the friction force a"= 1.96/2= 0.98ms^-2
Total acceleration of the body = a'+a" = 3.5+(-0.98)= 2.52ms^-2
the distance travelled by the body is given by the equation of motion = s= ut+1/2at^2
= 0+1/2×2.52×(10)^2= 126m
now remaining part of this question is given in picture..
=_==_==_==_==_==_==_==_==_==_==_==_==_==_=
I hope this answer help you
applied force= 7 N
coefficient of kinetic friction= 0.1
initial velocity/u= 0
time,t= 10sec.
The acceleration produced in the body by applied force is given by Newton's second law of motion as : acceleration(a)= f/m =7/2=
Now,
friction force given as: 0.1 ×2×9.8 =1.96N
the acceleration produced by the friction force a"= 1.96/2= 0.98ms^-2
Total acceleration of the body = a'+a" = 3.5+(-0.98)= 2.52ms^-2
the distance travelled by the body is given by the equation of motion = s= ut+1/2at^2
= 0+1/2×2.52×(10)^2= 126m
now remaining part of this question is given in picture..
=_==_==_==_==_==_==_==_==_==_==_==_==_==_=
I hope this answer help you
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