Question

3. A box of mass 15 kg is sliding along a smooth floor with a velocity of 15.0 ms. It
then enters a rough portion which has a length of 6.0 m. In this portion of the floor, a
frictional force of 80.0 N acts on the box. Determine (i) the work done by the frictional
force on the box? (ii) the velocity of the box when it leave the rough surface and (iii)
the length of the rough surface required to bring the box completely to rest. Take
8 = 10.0 ms-2​

Answer 1

Answer:

It then enters a rough portion which has a length of x=6.0 m. In this portion of the floor, a frictional force of F=80N acts on the box

Work done=F.x

=80\times680×6

=480J

b) frictional force

F=maF=ma

a=\frac{F}{m}=\frac{80}{15}=1.33m/s^2

m

F

=

15

80

=1.33m/s

2

Final velocity

v^2=u^2-2as\\=15^2-2\times1.33\times6v

2

=u

2

−2as

=15

2

−2×1.33×6

v=14.46m/s

c) for rest of box

v^2=u^2-2as\\0=15^2-2\times1.33\times sv

2

=u

2

−2as

0=15

2

−2×1.33×s

S=84.59m