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Question

A 5.82-kg object passes through the origin at time t = 0 such that its x component of velocity is 5.20 m/s and its y component of velocity is −3.12 m/s.

(a) What is the kinetic energy of the object at this time?

 ______ J

(b) At a later time t = 2.00 s, the particle is located at x = 8.50 m and y = 5.00 m. What constant force acted on the object during this time interval?

magnitude	____ N
direction	____ ° measured from the +x axis

(c) What is the speed of the particle at t = 2.00 s?

 _____m/s

Answer 1

Mass , m = 5.82 kg

vx(initial) = 5.2 m/s , vy(initial) = - 3.1 m/s

a)

Kinetic Energy , KE = 0.5 * m * vx^2 + 0.5 * m * vy^2

KE = 0.5 * 5.82 * ( 5.2^2 + 3.1^2) J

KE = 106.7 J

b)

at t = 2 s

x1 = 8.5 m , y1 = 5 m

Find the acceleration using the Position Equation

s(t) = s0 + v0 t + 1/2 a t^2

s(t) = v0 * t + 1/2 * a * t^2

8.5 i + 5 j = (5.2 i - 3.1 j) * 2 + 0.5 * a * ( 2^2)

a = (- 0.95 i + 5..6 j) m/s^2

The Force , F = m * a

F = 5.82 * (- 0.95 i + 5..6 j) N

F = (- 5.53 i +32.6 j) N

c)

the velocity , v = v0 + a * t

V = (5.2 i - 3.12 j) + (- 0.95 i + 5..6 j) * 2 m/s

v = 3.3 i m/s + 8.08 j m/s

the speed of particle , |v| = sqrt(3.3^2 + 8.08^2)

|v| = 8.73 m/s