Question
A 0.19 kg hockey puck has a velocity of 2.1 m/s toward the east (the +x direction) as it slides over the frictionless surface of an ice hockey rink. What are the (a) magnitude and (b) direction of the constant net force that must act on the puck during a 0.36 s time interval to change the puck's velocity to 4.1 m/s toward the west? What are the (c) magnitude and (d) direction if, instead, the velocity is changed to 4.1 m/s toward the south? Give your directions as positive (counterclockwise) angles measured from the +x direction.
i can easily find parts a and b but c and d i cant figure out the turn how to do it
i can easily find parts a and b but c and d i cant figure out the turn how to do it
Answers
a = (Vf-Vo)/t = (-4.1-2.1)/0.36 = -17.2 m/s^2.
a. F = M*a = 0.19 * (-17.2) = -3.27 N.
b. Direction = West.
c. a=(-4.1i-2.1)/0.36=(4.61[243o]/0.36
= 12.8 m/s^2[243o].
F = M*a = 0.19 * 12.8[243o] = 2.43N.[243o]
d. Direction = 243o CCW.
a. F = M*a = 0.19 * (-17.2) = -3.27 N.
b. Direction = West.
c. a=(-4.1i-2.1)/0.36=(4.61[243o]/0.36
= 12.8 m/s^2[243o].
F = M*a = 0.19 * 12.8[243o] = 2.43N.[243o]
d. Direction = 243o CCW.
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