A particle is located on the x axis at x = 2.0 m from the origin. A force of 25 N, directed 30° above the x axis in the x-y plane, acts on the particle. What is the torque about the origin on the particle?

25 Nm

Explanation:

Parameters given

Distance of the particle from the origin, r = 2.0 m

Force acting on particle, F = 25 N

Angle of force, θ = 30°

The torque acting on a particle is given as:

τ = r * F * sinθ

Where r = radius of axis of rotation. This is the same as the position of the particle on the x axis.

Therefore, the torque will be:

τ = 2 * 25 * sin30

τ = 25 Nm

The torque acting on the particle is 25 Nm.

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s