A particle, mass 0.25 kg is at a position () m, has a velocity () m/s, and is subject to a force () N. What is the magnitude of the torque on the particle about the origin

A particle, mass 0.25 kg is at a position (-7i + 7j + 5k) m, has a velocity (6i - j + 4k) m/s, and is subject to a force (-5i + 0j - k) N. What is the magnitude of the torque on the particle about the origin?

47.94Nm

The torque (τ) on a particle subject to a force (represented as force vector F) at a position (represented as position vector r) about the origin is given by the cross product of the position vector r for the point of application of a force and the force F. i.e

τ = r x F

Given:

r = (-7i + 7j + 5k) m

F = (-5i + 0j - k) N

                    |   i             j              k    |

r x F  =         |   -7            7              5  |

                    |  -5           0              -1   |

r x F  =       i(-7 - 0) - j(7+25) + k(0+35)

r x F  =       i(-7) - j(32) + k(35)

r x F  =       -7i - 32j + 35k

Therefore the torque τ = -7i - 32j + 35k

The magnitude of the torque is therefore;

|τ| =

|τ| =

|τ| =

|τ| = 47.94Nm

The magnitude of the torque on the particle about the origin is 47.94Nm

true

explanation:

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer