Answer to Question #240466 in Mechanics | Relativity for Anas

Question

Review, Three forces acting on an object are given by F1=

(-2.001 + 2.00j) N, and F2 = (5.001 - 3.00j) N, and F3=

(-45.01) N. The object experiences an acceleration of mag-

nitude 3.75 m/s?. (a) What is the direction of the accelera-

tion? (b) What is the mass of the object? (c) If the object is

initially at rest, what is its speed after 10.0 s? (d) What are

the velocity components of the object after 10.0 s?

Answer 1

Explanations & Calculations

• When forces act on a body, it may remain at rest or start moving uniform or at some acceleration. Here we are given the body to be moving at an uniform acceleration.
• If the body remains at rest, there is no any net force acting on the body & if it accelerates, it means there exist a net force. So here exist a net force due to the influence of those 3 forces.
• Finding the net, is easy when the forces are given in vector form similar to this case (otherwise, resolution of each force is to be considered). When given in vector form, the resultant/ the net force is easily obtained by the vector sum. This helps you for the first part.

$\qquad\qquad \begin{aligned} \small \vec{F_{net}}&=\small \vec{F_1}+\vec{F_2}++\vec{F_3}\\ &=\small (-2i+2j)+(5i-3j)+(-45i)\\ &=\small -42i-j\\\\ \small |\vec{F_{net}}|&=\small \sqrt{42^2+1}=1765N\\\\ \small \theta_{F_{net}}&=\small \tan^{-1}\Big(\frac{-1}{-42}\Big)\\ &=\small 1.36^0(South\,of\,West)(lies \,on\,the\,3^{rd}\,quadrant) \end{aligned}$

a)

• Now the magnitude & the direction of the net force are known. Direction of the acceleration always lie on the same direction as the net force. So you know the direction of the acceleration.

b)

• Since the net force & the acceleration under that force are known, it is now simply the usage of the Newton's second law (F=ma) to find the mass of the object. (Give it a try)

c)

• We can use the 4 motion equations ($\small v=u+at\,,\,v^2=u^2+2as\,,\,...)$ when the acceleration at which the object moves is constant. Now, here you are given a constant acceleration which means you need to use any appropriate equation to find the requested parameter.
• a, u & t are known & v after 10s is to be found. So $\small v=u+at$ isn't the appropriate equation to find the v. (Give it a try)

d)

• Now you know the direction the object moves in. It is the same as what you found with respect to the force & the acceleration $\small \theta.$ Velocity components can be defined with respect to the horizontal & the vertical.
• Since you now know the velocity, the horizontal & the vertical components are given as follows.

$\qquad\qquad \begin{aligned} \small v_{horizontal}&=\small v.\cos\theta\\ \small v_{verticle}&=\small v.\sin\theta \end{aligned}$

• Give this a try & ask on this site if you find any difficulty.