A catapult used to hold a stone of mass 500g is extended by 20cm with an applied force F.If the stone leaves with a velocity of 40ms-1,the value of F is
3 answers
400
X=vt
T=0.2/40=0.005s
A=v/t=40ms^-1/0.05s=8.0*10^3ms^-2
F=ma=0.5kg*8000ms^-2
=4.0*10^3N
T=0.2/40=0.005s
A=v/t=40ms^-1/0.05s=8.0*10^3ms^-2
F=ma=0.5kg*8000ms^-2
=4.0*10^3N
Ok go ahead
3 answers