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 40m/s, the value of F is
-
A.
4.0 x 102N -
B.
2.0 x 103N -
C.
4.0 x 103N -
D.
4.0 x 104N
Correct Answer: Option C
Explanation
Work done = kinetic energy of the stone
(frac{1}{2} F e = frac{1}{2} mv^2)
Given m = 500g = 0.5kg; v = 40m/s ; e = 20cm = 0.2m
(F = frac{mv^2}{e})
= (frac{0.5 times (40)^2}{0.2})
= 4.0 (times) 10(^3) N