Question

008 (part 1 of 3) 10.0 points A 0.338 kg particle has a speed of 3.8 m/s at point A and kinetic energy of 10.1 J at point B. What is its kinetic energy at A? Answer in units of J. 009 (part 2 of 3) 10.0 points What is the particle’s speed at B? Answer in units of m/s. 010 (part 3 of 3) 10.0 points What is the total work done on the particle as it moves from point A to B?

Answer 1

Answer:

1) 2.44 joules

2) 7.73 m/s

3) 7.6 joules

Explanation:

Kinetic energy (K) of a particle is:

$K=\frac{mv^{2}}{2}$ (1)

with m the mass, and v the velocity

1) Because we already now velocity on A (va) and the mass of the object we can calculate its kinetic energy:

$K_{a}=\frac{mv_{a}^{2}}{2}=\frac{(0.338kg)(3.8\frac{m}{s})^{2}}{2}=2.44J$

2) Because on B we know mass and kinetic energy we should solve (1) for v and use our values to find the velocity on B:

$v_{b}=\sqrt{\frac{2K_{b}}{m}}=\sqrt{\frac{2(10.1J)}{(0.338kg)}}=7.73\frac{m}{s}$

3) Work-energy theorem states that the change of kinetic energy of an object is equal to the total work done on it, so:

$W=K_b-K_a=10.1J-2.44J= 7.6J$