Answer :
a) mass m with F1 acting in the positive x direction and F2 acting perpendicular in the positive y direction
m = 5.00 kg
F1=20.0N ... x direction
F2=15.00N ... y direction
Net force ^2 = F1^2 + F2^2 = (20N)^2 + (15n)^2 = 625N^2 =>
Net force = √625 = 25N
F = m*a => a = F/m = 25.0 N /5.00 kg = 5 m/s^2
Answer: 5.00 m/s^2
b) mass m with F1 acting in the positive x direction and F2 acting on the object at 60 degrees above the horizontal.
m = 5.00 kg
F1=20.0N ... x direction
F2=15.00N ... 60 degress above x direction
Components of F2
F2,x = F2*cos(60) = 15N / 2 = 7.5N
F2, y = F2*sin(60) = 15N* 0.866 = 12.99 N ≈ 13 N
Total force in x = F1 + F2,x = 20.0 N + 7.5 N = 27.5 N
Total force in y = F2,y = 13.0 N
Net force^2 = (27.5N)^2 + (13.0N)^2 = 925.25 N^2 = Net force = √(925.25N^2) =
= 30.42N
a = F /m = 30.42 N / 5.00 kg = 6.08 m/s^2
Answer: 6.08 m/s^2
m = 5.00 kg
F1=20.0N ... x direction
F2=15.00N ... y direction
Net force ^2 = F1^2 + F2^2 = (20N)^2 + (15n)^2 = 625N^2 =>
Net force = √625 = 25N
F = m*a => a = F/m = 25.0 N /5.00 kg = 5 m/s^2
Answer: 5.00 m/s^2
b) mass m with F1 acting in the positive x direction and F2 acting on the object at 60 degrees above the horizontal.
m = 5.00 kg
F1=20.0N ... x direction
F2=15.00N ... 60 degress above x direction
Components of F2
F2,x = F2*cos(60) = 15N / 2 = 7.5N
F2, y = F2*sin(60) = 15N* 0.866 = 12.99 N ≈ 13 N
Total force in x = F1 + F2,x = 20.0 N + 7.5 N = 27.5 N
Total force in y = F2,y = 13.0 N
Net force^2 = (27.5N)^2 + (13.0N)^2 = 925.25 N^2 = Net force = √(925.25N^2) =
= 30.42N
a = F /m = 30.42 N / 5.00 kg = 6.08 m/s^2
Answer: 6.08 m/s^2
a) The acceleration of the block for the first configuration is [tex]\boxed{5\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^2}}}} \right.\kern-\nulldelimiterspace} {{{\text{s}}^2}}}}[/tex].
b) The acceleration of the block for the second configuration is [tex]\boxed{6.08\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}}[/tex].
Further Explanation:
Given:
The magnitude of force [tex]{F_1}[/tex] is [tex]20.0\,{\text{N}}[/tex].
The magnitude of force [tex]{F_2}[/tex] is [tex]15.0\,{\text{N}}[/tex].
The mass of the object is [tex]5.0\,{\text{kg}}[/tex].
Concept:
Part (a):
The situation in which the [tex]{F_1}[/tex] acts in positive X direction and [tex]{F_2}[/tex] in the positive Y direction is as shown in figure attached below.
Since the two forces are perpendicular to each other, the net force acting on the object will be:
[tex]\begin{aligned}{F_{net}}&=\sqrt{{{\left( {{F_1}} \right)}^2} + {{\left( {{F_2}}\right)}^2}}\\ &=\sqrt {{{\left( {20}\right)}^2} + {{\left( {15} \right)}^2}}\\ &= 25\,{\text{N}}\\\end{aligned}[/tex]
The angle at which the net force acts on the object is:
[tex]\begin{aligned}\theta&= {\tan ^{ - 1}}\left( {\frac{{{F_2}}}{{{F_1}}}} \right)\\&={\tan ^{ - 1}}\left({\frac{{15}}{{20}}} \right)\\&=36.86^\circ\\\end{aligned}[/tex]
The acceleration of the object under the action of the net force is given by:
[tex]\boxed{a=\dfrac{{{F_{net}}}}{m}}[/tex]
Substitute the values in above expression.
[tex]\begin{aligned}a&= \frac{{25}}{5}\,{{\text{m}}\mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}\\&= 5\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}\\\end{aligned}[/tex]
Thus, the acceleration of the object in the first configuration is [tex]\boxed{5\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}}[/tex] at an angle of [tex]36.86^\circ[/tex] from horizontal.
Part (b):
In the second configuration as shown in the figure attached below, the force [tex]{F_2}[/tex] is resolved into its horizontal and vertical components as follows:
[tex]\begin{aligned}{F_{2x}}&= {F_2}\cos 60\\&=7.5\,{\text{N}}\\\end{aligned}[/tex]
[tex]\begin{aligned}{F_{2y}}&= {F_2}\sin60\\&= 13\,{\text{N}}\\\end{aligned}[/tex]
The net force in the horizontal direction and the vertical direction are:
[tex]\begin{aligned}{F_x}&= {F_1} + {F_{2x}}\\&= 20 + 7.5\,{\text{N}}\\&= 27.5\,{\text{N}}\\\end{aligned}[/tex]
Thus, the resultant force acting on the object will be:
[tex]\begin{aligned}{F_{net}}&=\sqrt {{{\left({{F_x}} \right)}^2}+{{\left( {{F_y}} \right)}^2}}\\&=\sqrt {{{\left( {27.5}\right)}^2}+{{\left( {13} \right)}^2}}\\&= 30.42\,{\text{N}}\\\end{aligned}[/tex]
The angle at which the net force acts on the object is:
[tex]\begin{aligned}\theta&= {\tan ^{ - 1}}\left( {\frac{{{F_y}}}{{{F_x}}}}\right)\\&={\tan ^{ - 1}}\left({\frac{{13}}{{27.5}}} \right)\\&= 25.30^\circ\\\end{aligned}[/tex]
The acceleration of the object under the action of the net force is given by:
[tex]\boxed{a =\frac{{{F_{net}}}}{m}}[/tex]
Substitute the values in above expression.
[tex]\begin{aligned}a&= \frac{{30.42}}{5}\,{{\text{m}}\mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}\\&= 6.08\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}\\\end{aligned}[/tex]
Thus, the acceleration of the object in the first configuration is [tex]\boxed{6.08\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}}[/tex] at an angle of [tex]25.30^\circ[/tex] from horizontal.
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Answer Details:
Grade: High School
Subject: Physics
Chapter: Newton’s Law of Motion
Keywords:
Two forces, 5.0 kg object, acceleration of the object,F1=20 N, F2=15 N, for the configurations, net force, horizontal direction, vertical direction, perpendicular.
