Question

A particle of mass m = 0.3 kg is acted upon by a
force so that its position vector,

r(t)=2.0t^2(i)- 5.0t^2 (j) +t^2 (k)

where t is in s. What is the (instantaneous) power delivered by this force to the particle at time
t = 2.0s?

Answer 1

v = dr/dt = 4 t i - 10 t j + 2 t k

a = d^2r/dt^2 = 4 i - 10 j + 2 k

F = m a = .3 (4 i - 10 j + 2 k)

P = F dot v
= .3 (4 i - 10 j + 2 k)dot4 t i - 10 t j + 2 t k

=.3 (16+100+4)t = 36 t
at t = 2
P = 72 Watts