Question

a body of a mass 2 kg initally at rest moves uder the action of an applied horizental force of 7 N on a table woth coefficiant of kinetic friction 0.1 colmpute the work done by the applied force in 10s

Answer 1

FIRST U SHOULD WRITE ALL THE VALUES THAT ARE GIVEN THAT IS
M=2kg
U=0sec
F=7N
T=10sec
find the value of of work done

Answer 2

Fnet=F-uN where u is coefficient of friction

ma=7N-(0.1)2 x 10 N

=7N-2N=5N

a =5/2m/sec^2

now use kinematic formula

s=ut+1/2at^2

=0+1/2 (5/2)(10)^2=125m

a) workdone by applied force =force x displacement =7N x 125m=875 Nm

b) workdone by frictional = -f x S

= -2 x 125=-250Nm

c) workdone by net force =(2x 5/2N)x 125=625Nm

d) apply conservation of energy theorem

workdone by net force=change in kinetic energy =625Nm