Question

Two forces are acting on a 0.250 kg

hockey puck as it slides along the ice. The first force has a magnitude of 0.400 N
and points 25.0°
north of east. The second force has a magnitude of 0.540 N
and points 65.0
° north of east. If these are the only two forces acting on the puck, what will be the magnitude and direction of the puck's acceleration? Enter the direction as an angle measured in degrees counterclockwise from due east

Answer 1

a = 3.54 m/s²

direction = 48.1° north of east

Step-by-step explanation:

Add up the x and y components of the two applied forces, then determine the resultant, R

∑Fx = cos25(0.40 N) + cos65(0.54 N) = 0.591 N

∑Fy = sin25(0.40 N) + sin65(0.54 N) = 0.6585 N

R = √0.59² + 0.6585² = 0.884 N = Fnet

Ф = tan⁻¹(0.6585/0.59) = 48.1° north of east

Fnet = ma

a = Fnet/m = 0.884 N / (0.25 kg) = 3.54 m/s²