2. [In this question i and j are horizontal unit vectors.] A particle P of mass 5kg is moving on a smooth horizontal plane. The particle P is moving under the action of two forces (i+2j)N and (-ci-10j)N where c is a positive constant. The magnitude of the acceleration of P is 2ms^-2 Find the value of c

## Step 1: The motion of a particle is governed by Newton's second law, which states that the net force acting on a body equals its mass times its acceleration.#### Here, is the net force, is the mass of the particle and is the acceleration. ## Step 2:The net force , in this problem, is the vector sum of the two forces acting on the particle #### \(F=(F_1+F_2)\)where \(F_1= (\mathbf{i}+2 \mathbf{j})\,N\) and \(F_2=(-c \mathbf{i}-10 \mathbf{j}) \,N\).Substituting the given force vectors, we get:#### The force also equals to , where the mass and acceleration . So, our equation is easy to rewrite:#### ## Step3:We must find that the magnitude of the total force equals the magnitude of the mass times acceleration, comparing the magnitudes:A vector has magnitude \(√{((i_{component})^{2}+(j_{component})^{2})}, thus####| | = |\((5 * 2) \mathbf{i} + (– 10 \mathbf{j})\)|= \| (1-c)^2 + (-8)^2 \| = √(|\(((1 - c)² +. (-8)²)) = √(c^2 -2c - 63)By equating the magnitudes we get ### 10= √(c^2 -2c + 65)Squaring on both sides to get rid of the square root.### 100 = c^2 -2c + 65Rearranging this quadratic equation, we get:### c^2 - 2c - 35 = 0## Step4:Solve the above quadratic equation for , which is the roots of the equation. Using the quadratic formula \(x = (-b± √(b^2-4ac))/2a\), we can find the roots. So when a = 1, b = -2, c = -35.