a) The acceleration of the block for the first configuration is \(\boxed{5\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^2}}}} \right.\kern-
ulldelimiterspace} {{{\text{s}}^2}}}}\).
b) The acceleration of the block for the second configuration is \(\boxed{6.08\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-
ulldelimiterspace} {{{\text{s}}^{\text{2}}}}}}\).
Further Explanation:
Given:
The magnitude of force \({F_1}\) is \(20.0\,{\text{N}}\).
The magnitude of force \({F_2}\) is \(15.0\,{\text{N}}\).
The mass of the object is \(5.0\,{\text{kg}}\).
Concept:
Part (a):
The situation in which the \({F_1}\) acts in positive X direction and \({F_2}\) 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:
\(\begin{aligned}{F_{net}}&=\sqrt{{{\left( {{F_1}} \right)}^2} + {{\left( {{F_2}}\right)}^2}}\\ &=\sqrt {{{\left( {20}\right)}^2} + {{\left( {15} \right)}^2}}\\ &= 25\,{\text{N}}\\\end{aligned}\)
The angle at which the net force acts on the object is:
\(\begin{aligned}\theta&= {\tan ^{ - 1}}\left( {\frac{{{F_2}}}{{{F_1}}}} \right)\\&={\tan ^{ - 1}}\left({\frac{{15}}{{20}}} \right)\\&=36.86^\circ\\\end{aligned}\)
The acceleration of the object under the action of the net force is given by:
\(\boxed{a=\dfrac{{{F_{net}}}}{m}}\)
Substitute the values in above expression.
\(\begin{aligned}a&= \frac{{25}}{5}\,{{\text{m}}\mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}}\right.\kern-
ulldelimiterspace} {{{\text{s}}^{\text{2}}}}}\\&= 5\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-
ulldelimiterspace} {{{\text{s}}^{\text{2}}}}}\\\end{aligned}\)
Thus, the acceleration of the object in the first configuration is \(\boxed{5\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-
ulldelimiterspace} {{{\text{s}}^{\text{2}}}}}}\) at an angle of \(36.86^\circ\) from horizontal.
Part (b):
In the second configuration as shown in the figure attached below, the force \({F_2}\) is resolved into its horizontal and vertical components as follows:
\(\begin{aligned}{F_{2x}}&= {F_2}\cos 60\\&=7.5\,{\text{N}}\\\end{aligned}\)
\(\begin{aligned}{F_{2y}}&= {F_2}\sin60\\&= 13\,{\text{N}}\\\end{aligned}\)
The net force in the horizontal direction and the vertical direction are:
\(\begin{aligned}{F_x}&= {F_1} + {F_{2x}}\\&= 20 + 7.5\,{\text{N}}\\&= 27.5\,{\text{N}}\\\end{aligned}\)
Thus, the resultant force acting on the object will be:
\(\begin{aligned}{F_{net}}&=\sqrt {{{\left({{F_x}} \right)}^2}+{{\left( {{F_y}} \right)}^2}}\\&=\sqrt {{{\left( {27.5}\right)}^2}+{{\left( {13} \right)}^2}}\\&= 30.42\,{\text{N}}\\\end{aligned}\)
The angle at which the net force acts on the object is:
\(\begin{aligned}\theta&= {\tan ^{ - 1}}\left( {\frac{{{F_y}}}{{{F_x}}}}\right)\\&={\tan ^{ - 1}}\left({\frac{{13}}{{27.5}}} \right)\\&= 25.30^\circ\\\end{aligned}\)
The acceleration of the object under the action of the net force is given by:
\(\boxed{a =\frac{{{F_{net}}}}{m}}\)
Substitute the values in above expression.
\(\begin{aligned}a&= \frac{{30.42}}{5}\,{{\text{m}}\mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-
ulldelimiterspace} {{{\text{s}}^{\text{2}}}}}\\&= 6.08\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-
ulldelimiterspace}{{{\text{s}}^{\text{2}}}}}\\\end{aligned}\)
Thus, the acceleration of the object in the first configuration is \(\boxed{6.08\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-
ulldelimiterspace} {{{\text{s}}^{\text{2}}}}}}\) at an angle of \(25.30^\circ\) 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.
