Question

A force F (x) = (-5.0x2 + 7.0x) N acts on a particle. (a) How much work does the force do on the particle as it moves from x = 2.0 m to x = 5.0 m? (b) Picking a convenient reference point of the potential energy to be zero at x = co, find the potential energy for this force.

Answer 1

Answer:

Part a)

Work done is given as

W = -121.5 J

Part b)

Potential energy of the system is given as

$U  = - \frac{5x^3}{3} + \frac{7 x^2}{2}$

Explanation:

As we know that work done by variable force is given as

$W = \int F. dx$

here we know that

$F = - 5x^2 + 7 x$

so we have

$W = \int (-5 x^2 + 7x) dx$

so we have

$W = - 5x^3/3 + 7 x^2/2$

here we displace it from x = 2 to x = 5

so we will have

$W = -\frac{5}{3}(5^3 - 2^3) + \frac{7}{2}(5^2 - 2^2)$

$W = -195 + 73.5 = -121.5 J$

Part b)

As we know that change in Potential energy = work done

$U_f - U_i = -121.5$

$U - 0 = - \frac{5x^3}{3} + \frac{7 x^2}{2}$

$U  = - \frac{5x^3}{3} + \frac{7 x^2}{2}$

Answer 2

(a) The work done by the force is 121.5J

(b) The change in potential energy of the system is also 121.5J

Work and potential energy:

Given that the force acting on the particle is F(x) = (-5.0x²+7.0x)N

The particle travels from x = 2.0 m to x = 5.0 m.

(a) The work done is given by:

$W=\int\limits^a_b {F(x)} \, dx$

where a and b are the final and initial values of displacement x.

$W=-\int\limits^5_2 {(-5x^2+7x)} \, dx\\\\W=[5x^3/3-7x^2/2]_2^5$

W = 121.5J

(b) According to the work-energy theorem, the work done is equal to the change in potential energy of the system.

So the potential energy:

ΔPE = W

ΔPE = 121.5J

Learn more about work-energy theorem:

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