A horizontal force of 5 N acts on a body of
mass 2 kg initially at rest. It starts moving on
the table having coefficient of friction = 0.2.
Calculate:
(i) Work done by the applied force in 5 s.
(ii) Work done by force of friction in 5 s.
(iii) Work done by the net force in 5 s.
(iv) Change in kinetic energy of the body in 5 s.
What do you conclude from this example?
Answers
Answer:-
★ Given:-
Mass m = 3 kg
Force F = 8 N
Kinetic friction coefficient μ = 0.2
Initial velocity, u = 0
Time, t = 10 s
★ According to the Newton’s law of motion:-
= 2.6 m/s²
Friction force = μ mg
= 0.2 x 3 x -9.8 = -5.88 N
Acceleration due to friction:-
a"= = -1.96 m/s²
The total acceleration of the body = a′ + a”
= 2.6 + (-1.96) = 0.64 m/s²
★ According to the equation of the motion,
s =
=
= 32 m
★ Remember:-
(a) Wa = F x s = 8 x 32 = 256 J
(b) Wt = F x s = -5.88 x 32 = -188 J
Net force = 8 + (-5.88) = 2.12 N
(c) Wnet = 2.12 x 32 = 68 J
(d) Final velocity v = u + at
= 0 + 0.64 x 10 = 6.4 m/s
Change in kinetic energy = ½ mv² - ½ mu²
= ½ × 2 (v² - u²)
= 6.4² – 0²
= 41 J
Answer:
Answer:-
★ Given:-
Mass m = 3 kg
Force F = 8 N
Kinetic friction coefficient μ = 0.2
Initial velocity, u = 0
Time, t = 10 s
★ According to the Newton’s law of motion:-
{a}' ={\frac{F}{m}=\;{\frac{8}{3}}}a
′
=
m
F
=
3
8
= 2.6 m/s²
Friction force = μ mg
= 0.2 x 3 x -9.8 = -5.88 N
Acceleration due to friction:-
a"= \frac{-5.88}{3}
3
−5.88
= -1.96 m/s²
The total acceleration of the body = a′ + a”
= 2.6 + (-1.96) = 0.64 m/s²
★ According to the equation of the motion,
s = ut + \frac{1}{2} at^{2}ut+
2
1
at
2
= 0 + \frac{1}{2}\times\:0.64 \times{10}^{2}0+
2
1
×0.64×10
2
= 32 m
★ Remember:-
(a) Wa = F x s = 8 x 32 = 256 J
(b) Wt = F x s = -5.88 x 32 = -188 J
Net force = 8 + (-5.88) = 2.12 N
(c) Wnet = 2.12 x 32 = 68 J
(d) Final velocity v = u + at
= 0 + 0.64 x 10 = 6.4 m/s
Change in kinetic energy = ½ mv² - ½ mu²
= ½ × 2 (v² - u²)
= 6.4² – 0²
= 41 J