A body of mass 2Kg is resting on a rough horizontal surface.A force of 20N is mow applied to it for 10sec parallel to the surface.If the coefficient of kinetic friction between the surfaces in contact is 0.2,calculate
i) Work done by the applied force in 10sec.
ii) change in kunetic energy of the object in 10sec.
Answers
m = mass of the body = 2 kg
W = weight of the body = mg = 2 x 9.8 = 19.6 N
N = normal force on the body from surface below
Using equilibrium of force in vertical direction
N = W
N = 19.6 N
f = kinetic frictional force
μ = Coeffcient of kinetic friction = 0.2
kinetic frictional force is given as
f = μ N
f = (0.2) (19.6)
f = 3.92 N
F = applied force on the body = 20 N
a = acceleration of the body in horizontal direction
force equation in horizontal direction is given as
F - f = ma
20 - 3.92 = 2 a
a = 8.04 m/s²
t = time of travel = 10 sec
v₀ = initial velocity of the body = 0 m/s
d = displacement of the body in 10 sec = ?
Using the kinematics equation
d = v₀ t + (0.5) a t²
d = (0) (10) + (0.5) (8.04) (10)²
d = 402 m
work done by the applied force is given as
W = F d
W = 20 x 402
W = 8040 J
ii)
final velocity after 10 sec can be given as
v = v₀ + a t
v = 0 + (8.04) (10)
v = 80.4 m/s
change in kinetic energy is given as
ΔKE = (0.5) m (v² - v²₀)
ΔKE = (0.5) (2) ((80.4)² - 0²)
ΔKE = 6464.2 J
