We know that the formula to of acceleration is given by,
[tex]a=\frac{V_A}{t}[/tex]
Re-arrange to the Velocity,
[tex]V_B=at[/tex] (1)
We know as well for second Newton's law is,
[tex]F=ma[/tex]
We can re-arrange for the acceleration, so
[tex]a= \frac{F}{m}[/tex] (2)
Replacing (2) to (1),
[tex]V_B = \frac{F}{m}t[/tex]
Our values are,
[tex]F=6N[/tex]
[tex]m=3Kg[/tex]
[tex]t=1.5s[/tex]
[tex]V_A=4.5m/s[/tex]
Substituting the values,
[tex]V_B = \frac{6}{3}(1.5)[/tex]
[tex]V_B =3m/s[/tex]
We have two components of a net velocity, so we can calculate the direct force through,
[tex]V= \sqrt{V_A^2+V_B^2}[/tex]
[tex]V=\sqrt{(4.5)^2+(3)^2}[/tex]
[tex]V=5.40m/s[/tex]
Answer:
The northward component of the puck’s velocity is 3.0 m/s.
The speed of puck is 5.40 m/s.
Explanation:
Given data:
Mass of puck is, [tex]m=3.0 \;\rm kg[/tex].
Speed at East is, [tex]v_{e}=4.5 \;\rm m/s[/tex].
Magnitude of force at north is, [tex]F = 6.0 \;\rm N[/tex].
Time interval is, [tex]t= 1.5 \;\rm s[/tex].
(a)
The north component of velocity is obtained by expression of force as,
[tex]f = ma\\f= m \times \dfrac{v_{n}}{t} \\6.0= 3 \times \dfrac{v_{n}}{1.5}\\v_{n}=3 \;\rm m/s[/tex]
Thus, the northward component of the puck’s velocity is 3.0 m/s.
(b)
The speed of puck is,
[tex]v'=\sqrt{v^2_{e}+v^2_{n}} \\v'=\sqrt{4.5^2+3.0^2} \\v'=5.40 \;\rm m/s[/tex]
Thus, the speed of puck is 5.40 m/s.
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