Question

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 following and interpret your resultsa. 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 sd. Change in kinetic energy of the body in 10 s.

Answer 1

natak energy of the body is 10 second

Answer 2

Hii dear,

● Answer-
a) Work done by applied force = 875 J
b) Work done by friction = -250 J
c) Work done by net force = 625 J
d) Change in KE = 625 J

● Explainations-
# Given-
m = 2 kg
F = 7 N
μ = 0.1

# Solution-
Here, Force of friction is
f = μmg
f = 0.1×2×10
f = 2 N

Net force is
Fnet = F-f
Fnet = 7-2
Fnet = 5 N

Acceleration here is calculated by
a = Fnet/m
a = 5/2
a = 2.5 m/s

Displacement of body,
s = 1/2 at^2
s = 0.5×2.5×(10)^2
s = 125 m

a) Work done by applied force,
W = F.s
W = 7×125
W = 875 J

b) Work done by frictional force,
W = -f.s
W = -2×125
W = -250 J

c) Work done by net force,
W = Fnet.s
W = 5×125
W = 625 J

d) Change in K.E.
KEf - KEi = Work
KE(change) = 625 J

Hope that is useful...