Asked by Mark
Two forces are applied to a 0.5 kg hockey puck causing it to accelerate. The first force has a magnitude of 10.0 N and is applied to the puck at an angle of 80 degrees with respect to the x axis. The second force has a magnitude of 3.5 N and is applied to the puck at an angle of 10 degrees with respect to the x axis.
What is the magnitude of the acceleration?
What is the magnitude of the acceleration?
Answers
Answered by
Henry
Fr = F1+F2 = 10N[80o] + 3.5N[10o].
X = 10*Cos80 + 3.5*Cos10 = 5.18 N.
Y = 10*sin80 + 3.5*sin10 = 10.5 N.
Fr = = sqrt(x^2+y^2) = (5.18^2+10.5^2) = 11.7 N, = Resultant force.
Fr = M*a = 11.7, a = 11.7/M = 11.7/0.5 = 22.14 m/s^2.
X = 10*Cos80 + 3.5*Cos10 = 5.18 N.
Y = 10*sin80 + 3.5*sin10 = 10.5 N.
Fr = = sqrt(x^2+y^2) = (5.18^2+10.5^2) = 11.7 N, = Resultant force.
Fr = M*a = 11.7, a = 11.7/M = 11.7/0.5 = 22.14 m/s^2.
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