Question

Two forces F1 and F2 act on a 5.00 kg object. Taking F1=20.0N and F2=15.00N, find the acceleration of the object for the configurations:
a) mass m with F1 acting in the positive x direction and F2 acting perpendicular in the positive y direction

b) mass m with F1 acting in the positive x direction and F2 acting on the object at 60 degrees above the horizontal.

Answer 1

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

Answer 2

a) The acceleration of the block for the first configuration is $\boxed{5\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^2}}}} \right.\kern-\nulldelimiterspace} {{{\text{s}}^2}}}}$.

b) The acceleration of the block for the second configuration is  $\boxed{6.08\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}}$.

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.

$\begin{aligned}a&= \frac{{25}}{5}\,{{\text{m}}\mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}\\&= 5\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}\\\end{aligned}$

Thus, the acceleration of the object in the first configuration is $\boxed{5\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}}$ 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.

$\begin{aligned}a&= \frac{{30.42}}{5}\,{{\text{m}}\mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}\\&= 6.08\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}\\\end{aligned}$

Thus, the acceleration of the object in the first configuration is $\boxed{6.08\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}}$ at an angle of $25.30^\circ$ from horizontal.

Learn More:

1. A 30.0-kg box is being pulled across a carpeted floor by a horizontal force of 230 N brainly.com/question/7031524

2. Choose the 200 kg refrigerator. Set the applied force to 400 n (to the right). Be sure friction is turned off brainly.com/question/4033012

3. Which of the following is not a component of a lever brainly.com/question/1073452

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.