Question

A 0.41-kg particle has a speed of 5.0 m/s at point A and kinetic energy of 8.5 J at point B. (a) What is its kinetic energy at A? J (b) What is its speed at point B? m/s (c) What is the total work done on the particle as it moves from A to B?

Answer 1

Answer:

A)The kinetic energy at a point A when a particle of .41 kg moves at a speed of 5.0 m/s is 5.125 J.

B)The speed at point B of the particle of .41 kg with Kinetic Energy 8.5 J is 6.44 m/s.

C)The total Work done by the particle is 3.375 J.

A)Explanation

From the given statements, we know  

Mass (m) = 0.41 kg

Velocity = 5.0 m/s

As we know Kinetic Energy (KE) = $\frac{1}{2} \text { mass } \times \text { velocity }^{2}$

Substituting the values in the above equation, we find

Kinetic Energy (KE) = $\frac{1}{2} \text { mass } \times \text { velocity }^{2}$

$\frac{1}{2} \times 0.41 \times 5^{2}$= 5.125 joule

(b) Explanation

From the given statements, we know  

Mass (m) = 0.41 kg

Kinetic Energy (KE) = 8.5 J

As we know Kinetic Energy (KE) = $\frac{1}{2} \text { mass } \times \text { velocity }^{2}$

Substituting the values in the above equation, we find

Kinetic Energy (KE) = $\frac{1}{2} \times 0.41 \times \mathrm{v}^{2}$ = 8.5 J

Or, $v^2$ = $\frac{8.5 \times 2}{0.41}$ = $\frac{8.5 \times 2}{0.41}$

Or, v = $\sqrt{\frac{17}{0.41}}$=6.439 m/s ~ 6.44 m/s

(c) Explanation

From the above given statements, we know  

Kinetic Energy at A = 5.125 J

Kinetic Energy at B = 8.5 J

As we know

Work Done = Change in Kinetic Energy (ΔKE)

So, Work Done (WD) = Kinetic Energy at B - Kinetic Energy at A

  WD = (8.5 – 5.125) J = 3.375 J