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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

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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

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

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GraceVeteran · Tutor for 9 years

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\(c = 7 ; c= -5\), but since c must be positive, then final value of \(c\) is 7.

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## Step 1: <br /><br />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.<br /><br />#### \(F = m*a\)<br /><br />Here, \(F\) is the net force, \(m\) is the mass of the particle and \(a\) is the acceleration. <br /><br />## Step 2:<br /><br />The net force \(F\), in this problem, is the vector sum of the two forces acting on the particle <br /><br />#### \(F=(F_1+F_2)\)<br /><br />where \(F_1= (\mathbf{i}+2 \mathbf{j})\,N\) and \(F_2=(-c \mathbf{i}-10 \mathbf{j}) \,N\).<br /><br />Substituting the given force vectors, we get:<br /><br />####\[F=(\mathbf{i}+2 \mathbf{j})+(-c \mathbf{i}-10 \mathbf{j}) = (1-c) \mathbf{i}+ (- 8) \mathbf{j}\,N \]<br /><br />The force \(F\) also equals to \(m*a\), where the mass \(m=5 kg\) and acceleration \(a = 2 \, m/s^2\). So, our equation is easy to rewrite:<br /><br />#### \(F = m * a = 5 * 2 = 10 \, N\)<br /><br />## Step3:<br /><br />We must find that the magnitude of the total force equals the magnitude of the mass times acceleration, comparing the magnitudes:<br /><br />A vector has magnitude \(√{((i_{component})^{2}+(j_{component})^{2})}, thus<br /><br />####|\(F\)| = |\((5 * 2) \mathbf{i} + (– 10 \mathbf{j})\)|= \| (1-c)^2 + (-8)^2 \| = √(|\(((1 - c)² +. (-8)²)) = √(c^2 -2c - 63)<br /><br />By equating the magnitudes we get <br /><br />### 10= √(c^2 -2c + 65)<br /><br />Squaring on both sides to get rid of the square root.<br /><br />### 100 = c^2 -2c + 65<br /><br />Rearranging this quadratic equation, we get:<br /><br />### c^2 - 2c - 35 = 0<br /><br />## Step4:<br /><br />Solve the above quadratic equation for \(c\), 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.
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