Help me solve this physics problem Two forces are acting on a 0.150-kg hockey pu

Question

Help me solve this physics problem

Two forces are acting on a 0.150-kg hockey puck as it slides along the ice. The first force has a magnitude of 0.345 N and points 25.0degree north of east. The second force has a magnitude of 0.525 N and points 55.0degree 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. magnitude of acceleration = direction of acceleration =

Explanation / Answer

vector representation of force 1 = F1 = 0.345*cos25 i + 0.345*sin25 j = ( 0.313 i + 0.146 j ) N

vector representation of force 2 = F2 = 0.525*cos55 i + 0.525*sin55 j = ( 0.301 i + 0.430 j ) N

Net force on object F = F1 + F2 = ( 0.313 i + 0.146 j ) + ( 0.301 i + 0.430 j ) = ( 0.614 i + 0.576 j ) N

magnitude of F = sqrt(0.614^2 + 0.576^2) = 0.842 N

direction = arctan(0.576/0.614) = 43.17 degrees counterclockwise from due east

so magnitude of acceleration = F/m = 0.842/0.15 = 5.61 m/s^2

direction of acceleration = direction of force = 43.17 degrees counterclockwise from due east

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