Question

A 6.22-kg object passes through the origin at time t = 0 such that its x component of velocity is 5.10 m/s and its ycomponent of velocity is -3.27 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

Answer 1

a) KE = 1/2 m v^2

magnitude of velocity = Sqrt[5.10^2 +(-3.27)^2] so that v^2 = 5.35^2 + 2.94^2 = 37.3 m^2/s^2 so v = 6.05 m/s

and KE = 1/2 m v^2 = 1/2 x 6.22 kg x v^2 = 113.97J

b) we will find the force by finding the acceleration

dist traveled = v0 t + 1/2 a t^2

to find x component of accel:

x dist = v0x t + 1/2 ax t^2

8.5 m = 5.10m/s x 2s + 1/2 ax (2s)^2

ax = -0.85m/s/s

for ay:

5m = -3.27m/s x 2s + 1/2 ay (2s)^2

ay = 5.77m/s/s

total accel = Sqrt[ax^2 + ay^2] = 5.83 and F = m a

the velocity at t=2s will be

vx = v0x + ax * 2s = 5.10m/s -0.85m/s/s x 2s = 3.40m/s

vy = v0y + ay x 2s = -3.27m/s + 5.77m/s/s x 2s = 8.27 m/s