Question

velocity time graph of a block of mass 100 gram sliding on horizontal concrete floor under the action of a constant force of 5 N shown below the magnitude of frictional Force acting in the block due to the floor is

Answer 1

From the velocity time graph we can note:- Final velocity= 40m/s Initial velocity= 20m/s Time= 2s {4s-2s= 2s} Acceleration= (v-u)/t                      = (40m/s -20m/s) ÷2s                      = 20m/s ÷2s                      = 10m/s^2 Mass of the block= 100g=100/1000kg                               = 0.1 kg Force= mass × acceleration Force= 0.1 kg × 10m/s^2 Force= 1 N Constant force - Frictional force= Net force 5N- Frictional force= 1N -Frictional Force= 1N - 5N - Frictional Force = -4N Frictional Force= 4N

Answer 2

from the graph

u= 60m/s

v=80m/s

t= 8-6=2sec.

from 1st eq. of motion

v=u+at

80=60+a(2)

80-60=2a

20=2a

20/2=a

10m/s^2= a

mass = 100gram = 0.1kg

force=ma=0.1×10=1N

constant force =5N(given)

constant force - frictional force = net force

let frictional force be F

5-F=1

5-1=F

4=F

frictional force = 4N

hope it helps u