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?

By: |Published on: Nov 16|Categories: Physics JAMB| 0 comments

A. 4.0 x 102N
B. 2.0 x 103N
C. 4.0 x 103N
D. 4.0 x 104N

Correct Answer:

Option C = 4.0 x 103N

Explanation

Work done = kinetic energy of the stone

½Fe= ½mv²
Given m = 500g = 0.5kg; v = 40m/s ; e = 20cm = 0.2m

F=mv²/e
= 0.5×(40)²/0.2

= 4.0 × 103 N

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