Question

A bullet of mass 0.04kg moving with a speed of 90m/s enters a heavy wooden block and is stopped after a distance of 60 cm.What is the average resistive force exerted by the block on the bullet?​

Answer 1

Answer:

Explanation:

Mass= 0.04kg

Vel = 90

Work done = 1/2 m v^2

= 1/2x 0.04 x 8100

= 162 joule

W= 162

Distance = 0.6 metre

F=?

W= fxd

162= fx 0.6

F= 162/0.6

=270N

Answer 2

Answer:

270N

Explanation:

MASS =0.04KG

INTITIAL SPEED(U)=90M/S

FINAL SPEED(V)=0M/S

DISTANCE TRAVELLED =0.6M

SO BY EQUATION

V^2-U^2=-2AS

=0-(90)^2=-2×A×0.6

=A=90×15×5

NOW

FORCE = MASS×ACCLERATION

= 0.04×90×15×5

=270N