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
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option C
Explanation:
Work done = kinetic energy of the stoneÂ
\(\frac{1}{2} F e = \frac{1}{2} mv^2\)
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
Work done = kinetic energy of the stoneÂ
\(\frac{1}{2} F e = \frac{1}{2} mv^2\)
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