Question

A 3.0 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

Answer 1

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

$a=\frac{V_A}{t}$

Re-arrange to the Velocity,

$V_B=at$ (1)

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

$F=ma$

We can re-arrange for the acceleration, so

$a= \frac{F}{m}$ (2)

Replacing (2) to (1),

$V_B = \frac{F}{m}t$

Our values are,

$F=6N$

$m=3Kg$

$t=1.5s$

$V_A=4.5m/s$

Substituting the values,

$V_B = \frac{6}{3}(1.5)$

$V_B =3m/s$

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

$V= \sqrt{V_A^2+V_B^2}$

$V=\sqrt{(4.5)^2+(3)^2}$

$V=5.40m/s$