The body was moving at a speed v1 = 3m / s. A constant force F = 4H was applied to the body during

| September 4, 2021 | education

The body was moving at a speed v1 = 3m / s. A constant force F = 4H was applied to the body during delta t = 2 s. During this time, the kinetic energy of the body has increased by delta Ek = 100J. Determine: 1) m 2) V2 – the speed of the body at the end of the action of the force.

Data: V1 (initial speed) = 3 m / s; t2 (duration of the force) = 2 s; F (constant force) = 4 N; ΔEk (change in kinetic energy) = 100 J.

1) Find the body mass: ΔEk = m * V2 ^ 2/2 – m * V1 ^ 2/2 = m * (V1 + a * t) 2/2 – m * V1 ^ 2/2 = m * (V1 + F / m * t) ^ 2/2 – m * V1 ^ 2/2 = m * (3 + 4 / m * 2) ^ 2/2 – m * 3 ^ 2/2 = m * (3 + 8 / m) ^ 2/2 – 4.5m = m * (9 + 48 / m + 64 / m ^ 2) / 2 – 4.5m = 4.5m + 24 + 32 / m – 4.5m = 24 + 32 / m = 100 J.

32 / m = 100 – 24 = 76.

m = 32/76 = 0.421 kg.

2) Determine the speed of the body after 2 seconds of the action of the force: V2 = V1 + a * t = V1 + F / m * t = 3 + 4 / 0.421 * 2 = 22 m / s.

Answer: Body weight is 0.421 kg; after 2 s, the body will move at a speed of 22 m / s.



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