A3.0 kg puck slides due east on a horizontal frictionless sur-face at a constant speed of 4.5 m/s. then a force of magnitude 6.0 n, directed due north, is applied for 1.5 s. afterward, a. what is the northward component of the puck’s velocity? a. 0.50 m/s d. 4.0 m/s b. 2.0 m/s e. 4.5 m/s b. what is the speed of the puck? a. 4.9 m/s d. 7.5 m/s b. 5.4 m/s e. 11 m/s c. 3.0 m/s c. 6.2 m/s

We know that the formula to of acceleration is given by,

Re-arrange to the Velocity,

(1)

We know as well for second Newton's law is,

We can re-arrange for the acceleration, so

(2)

Replacing (2) to (1),

Our values are,

Substituting the values,

We have two components of a net velocity, so we can calculate the direct force through,

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