Question

a particle of mass 0.1 kg is subjected to a force which varies with distance as shown in figure if it starts its journey from rest at x=0 its velocity at x=12m will be

Answer 1

Answer:

v = 40m/s

Given:

m = 0.1 kg

Solution:  

Area under the graph can be computed by,

IMAGE

m = 0.1 kg  

For area (0-4) = ½ (upper limit – lower limit) x F(N) = ½ (4-0) x (10) = 20

For area (4-8) = ½ (upper limit – lower limit) x F(N) = (8-4) x (10) = 40

For area (8-12) = ½ (upper limit – lower limit) x F(N) = ½ (12-8) x (10) = 20

Total area = 20+40+20 = 80

To find velocity =1/2 m x $v^2$

1/2 m x $v^2$  = 80

(0.1) x $v^2$ = 160

$v^2$= 160/0.1 = 1600

v = √1600 = 40m/s

Answer 2

Answer:

40 m/s

Explanation:

w= force *distance

so workdone = area under F-x graph