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5 a particle pof mass m=0.3kg moves under the action of a single constant force f newtons. the acceleration of p is a=(5i+7j)ms^-2 (2

Question

5 A particle Pof mass m=0.3kg moves under the action of a single constant force F newtons. The acceleration of P is a=(5i+7j)ms^-2 (2 marks) a Find the angle between the acceleration and i. Force, mass and acceleration are related by the formula F=ma b Find the magnitude of F. (3 marks)

Answer

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Ina Master · Tutor for 5 years

Answer

a)The angle θ between the force vector F and the unit vector i is obtained through considering that cos(θ) = (F.i)/|F| = (F.i)/(sqrt(F.i^2 + F.j^2)). Which results in cos(θ) = 5/sqrt(5^2 + 7^2) = 5/sqrt(74). Therefore, θ = arcos(5/sqrt(74)), which is the angle in radians. If you wish to convert this into pounds, multiply by 180/π.b) Now to find the magnitude of , we compute the equation F=m|a| which means plugging in given values F= (0.3 sqrt (5 ^ 2 + 7^ 2) = 156 Newtons

Explanation

This question explores fundamental principles of physics, primarily the concept of force, mass, and acceleration. The principle, often named after Isaac Newton, is understood as: Force (F) = Mass (m) * Acceleration (a). For part (a) of the question, we are required to find the angle between the acceleration vector and i (`i` being the unit vector along the x-axis). This can be done using the concept of dot products. Also, as we know that, the dot product of a vector (in this case, acceleration) and the unit vector along a given direction gives us the component of the vector in that direction. For part (b) of the question, we need to find the magnitude of the vector F that represents the force given that F = maWhen we want to find the magnitude of a force, we use the formula ^^F = m*|a|^, where m is mass and |a| is the magnitude of acceleration.where |a| = sqrt(a.i^2 + a.j^2)