Question

Two forces are acting on a 0.150-kg hockey puck as it slides along the ice. The first force has a magnitude of 0.405 N and points 35.0° north of east. The second force has a magnitude of 0.605 N and points 55.0° 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. A) Magnitude of acceleration = _____ m/s2 B) Direction of acceleration = ____ degrees

Answer 1

Fx = 0.405 * cos(35.0) + 0.605 * cos(55.0) N
Fx = 0.679 N

Fy = 0.405 * sin(35.0) + 0.605 * sin(55.0) N
Fy = 0.728 N

ax = Fx/m
ax = 0.679/0.150
ax = 4.53 m/s^2

ay = Fy/m
ay = 0.728/0.150
ay = 4.85 m/s^2

Net Magnitude of Acceleration, a = sqrt(ax^2 + ay^2)
a = sqrt(4.53^2 + 4.85^2)
a = 6.64 m/s^2

Direction = tan^-1(4.85/4.53)
Direction = 47.0^o Counterclockwise from due east.