An object of mass 3.9 kg is moving in a straight line with kinetic energy 137.59

Question

An object of mass 3.9 kg is moving in a straight line with kinetic energy 137.592 J. A force is applied in the direction of its motion for 3 seconds, and as a result, its kinetic energy is multiplied by a factor of 1.51. a) By what factor is its momentum multiplied? b) What was the magnitude of the force applied to the object? N c) Instead of this force, if a force of the same magnitude is applied to the object for the same duration, but is perpendicular to the direction in which the object was originally moving, by what factor is the kinetic energy multiplied?

Explanation / Answer

a) we know,

KE = (1/2)*m*v^2

= (m*v)^2/(2*m)

KE = P^2/(2*m)

==> P = sqrt(2*KE*m)

so, Pi = sqrt(2*m*KEi)

Pf = sqrt(2*m*KEf)

Pf/Pi = sqrt(KEf/KEi)

Pf = Pi*sqrt(1.51*KEi/KEi)

Pf = 1.23*Pi

momentum is multiplied by 1.23

b) use Impulse-Momtnum theorem,

F*t = Pf - Pi

F = (Pf -Pi)/t

= (1.51*KEi - KEi)/t

= 0.51*137.592/3

= 23.4 N

c) same. 1.51

because, workdone is same in both cases.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.