Question

A single conservative force acts on a 5.00-kg particle. The equation N, where x is in meters, describes this force. As the particle moves along the x axis from m to m, calculate (a) the work done by this force, (b) the change in the potential energy of the system, and (c) the kinetic energy of the particle at m if its speed at m is 3.00 m/s.

Answer 1

Answer:

a) $W=69.44 J$

b) $\Delta U=-69.44 J$

c) $K_{f}=91.94 J$

Explanation:

a) The work is defined as the integral of the force from x₁ to x₂:

$W=\int^{x_{2}}_{x_{1}}Fdx$

Here:

• F is the force F=2x+4
• x₁ is 1.4 m
• x₂ is 7.0 m

Now. we need to take the integral to get the work done:

$W=\int^{x_{2}}_{x_{1}}(2x+4)dx$

$W=x^{2}|^{x_{2}}_{x_{1}}+4x|^{x_{2}}_{x_{1}}$

$W=(7^{2}-1.4^{2})+4(7-1.4)=69.44 J$

b) The change of the potential energy is minus de work, so:

$\Delta U=-\Delta W$

$\Delta U=-69.44 J$

c) We know that the change in kinetic energy is equal to the work done, in a conservative system, so:

$W=\Delta K= 69.44 J$

And know that:

$\Delta K= K_{f}-K_{i}$

We need to find the final kinetic energy, so let's solve it for Kf

$K_{f}=\Delta K+K_{i}$

$K_{f}=\Delta K+(1/2)mv^{2}$

• m is 5 kg
• v is 3 m/s

$K_{f}=69.44+(1/2)*5*3^{2}$

$K_{f}=91.94 J$

I hope it helps you!