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This is one of our many quizzes on Newton's first law. Click on this to access more such competitive quizzes on Dynamics.

Newton's first law Quiz 2

physics

 Assume that a football is at rest and its state of equilibrium cannot be changed until an external force is applied on it. With this regard, which of the following statements would you consider to be true?

 Consider an object moving with a certain velocity. It won’t stop itself unless it is compelled by a certain force. The same applies to the object placed at rest. You can also consider the law of inertia. 

  $\text{According to Newton’s first law of motion,} \\ \text{a body will remain at rest or it will keep moving in a straight line at a constant speed,} \\ \text{unless, a force acts upon it}\text{.} \\ \text{It is also known as the law of inertia}\text{.} \\ \text{You can change the state of motion or rest by applying some net force} \\ \text{it acts in accordance to the velocity of the object}\text{.}$

Suppose you are riding on a giant wheel, on which the net force of the cart acts downward and the cart itself is moving towards right. If the speed of the cart is neither decreasing or increasing at a certain rate, then determine the path of the cart. <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/4e864d1f-5041-4987-9c7e-11b94d4a11547769961875108361546.png"/>

 Assume that net force acting downwards has a turning point due to which a trajectory is generated that provides for a constant acceleration along the direction of net force.

  $\text{While riding on the wheel, the cart exerts a downward net force,} \\ \text{and, itself it moves towards right direction}\text{. } \\ \text{Also, the speed of the cart is neither increasing nor decreasing}\text{.} \\ \text{It is because of the reason that the cart follows a curved trajectory path,} \\\text{where, net force is perpendicular to the velocity vector}\text{.} \\ \text{Also the net force and acceleration are in the same direction}\text{. } \\ \text{Thus, we can say that the }{{F}_{net}}\text{ has a turning point that leads to a curved path}\text{.}$

Suppose a ball of $75g$ rests on ramp ‘${R_1}$’ making an angle of ${{45}^{\circ }}$ and bears its weight on the vertical ramp ‘${R_2}$’. What would be the value of surface reactions at ramps ${R_1}$ and ${R_2}$ as shown in the figure? <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/ce2cea98-0eef-48c8-8b97-16ed7b410b853280479983636935580.png"/>

 Consider the surface in contact with the ball that will produce the surface reaction forces using the free body diagram. Also by rotating the axis you will find the components of force with respect to x and y-axis.

&nbsp;$\text{In order to find the surface reactions,} \\ \text{first we need to calculate the total forces acting on x and y-axis}\text{.} \\ \text{Using the concept of rotation of axis,} \\ \text{we have one sin component on x-axis,} \\ \text{and one cos component on y-axis}\text{.} \\ \text{Now, taking into consideration the equilibrium condition,} \\ \text{we get:} \\ \sum{{{F}_{x}}=0} \\ \therefore -{{\text{N}}_{\text{2}}}\text{+}{{\text{N}}_{1}}\text{sin4}{{\text{5}}^{\circ }}\text{=0} \\ \text{or, }{{\text{N}}_{1}}\text{sin4}{{\text{5}}^{\circ }}={{\text{N}}_{\text{2}}} \\ \Rightarrow {{\text{N}}_{1}}=\sqrt{2}{{\text{N}}_{\text{2}}}\text{ }...\text{(i)} \\ \text{and for y-axis, we get:} \\ \sum{{{F}_{y}}=0} \\ \therefore {{\text{N}}_{1}}\text{cos4}{{\text{5}}^{\circ }}\text{-75=0} \\ \text{on converting }75g\text{ into newton, } \\ \text{we get }75\times 10=750N \\ \text{Thus, }{{\text{N}}_{1}}\text{cos4}{{\text{5}}^{\circ }}\text{-750N=0} \\ \text{or, }{{\text{N}}_{1}}\times \dfrac{1}{\sqrt{2}}=500N \\ \Rightarrow {{\text{N}}_{1}}=500\sqrt{2}N\text{ }...\text{(ii)} \\ \text{Thus, on comparing equation (i) with (ii), we get:} \\ \therefore {{\text{N}}_{\text{2}}}=500\sqrt{2}.\sqrt{2}=1000N \\ \text{as }{{\text{N}}_{\text{2}}}\text{ is the actual contact force of the surface with ball}\text{.}$ <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/630e235a-792b-44ee-80fa-b2ae4e044c693197338480828744110.png"/>

  A cubical box after losing its balance slides down a ${{52}^{\circ }}$ ramp that has a coefficient of kinetic friction of $0.37$. What is the acceleration of the box? 

 Recall the concept of rotation of axis to assume the sin and cos components of the box placed on the ramp. You can also take the help of the free body diagram to get to the solution.

$\text{Let us consider the force down the ramp be:} \\ mg\sin \theta \\ \text{and the force due to kinetic friction pointing up the ramp:} \\ F={{\mu }_{k}}.{{F}_{n}} \\ \text{where, }{{\mu }_{k}}\text{ is the coefficient of kinetic friction} \\ {{F}_{n}}\text{ is the normal force} \\ \text{Also, }{{F}_{n}}=m.g.\sin \theta \\ \text{and, the force due to kinetic friction is:} \\ {{F}_{n}}={{\mu }_{k}}.m.g.\cos \theta \\ \text{Thus, the net force on the box along the ramp is given by:} \\ \Rightarrow F=m.g.\sin \theta -{{\mu }_{k}}.m.g.\cos \theta \\ \text{On dividing the above equation by }\!\!'\!\!\text{ }m\text{ }\!\!'\!\!\text{ ,} \\ \text{we get:} \\ \text{acceleration, a}=g.\sin \theta -{{\mu }_{k}}.g.\cos \theta \\ \text{on substituting the values, we get:} \\ \text{a}=9.8\times 0.78-0.37\times 9.8\times 0.62 \\ \text{a}=5.4m/{{\sec }^{2}}$ <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/d8fe5861-fab3-4c91-bb69-c687703c8c252373123613162785829.png"/>

 A boy stands on a wooden ramp elevated at an angle of ${{30}^{\circ }}$ to the horizontal.The magnitude of mass of the boy is $64\,Kg$. Find out the force exerted by the normal reaction of the surface on mass?

 Consider the free body diagram to find the axis of origin of normal reaction that exerts a force on the surface of mass. By calculating the forces of the respective axis we will get the normal force acting on its surface.

$\text{Using the concept of rotation of axis,} \\ \text{we can find the cos and sin components of both x and y-axis}\text{.} \\ \text{Using these sin and cos components we can find the forces along the x and y-axis}\text{.} \\ \text{Given, }\theta \text{=3}{{\text{0}}^{\circ }} \\ \text{Now, force of gravity exerted by the boy is:} \\ Fg=mg, \\ \text{where, }m=64kg\text{ (given) and }g=9.8N \\ \therefore Fg=64\times 9.8=627.2N \\ \text{In order to find the normal reaction force, } \\ \text{we will consider only y-axis as the normal forces act on only y-axis}\text{.} \\ \text{On y-axis, }-Fg\text{cos3}{{\text{0}}^{\circ }}\text{ component lies} \\ \therefore Fgy=-Fg\cos {{30}^{\circ }} \\ \Rightarrow -627.2\times 0.1543=-96.78N \\ \text{Assuming the total forces of }Fy\text{ to be} \\ \sum{{{F}_{y}}={{F}_{N}}+Fgy,} \\ \text{where, }{{F}_{N}}\text{ is the normal force acting on the surface}\text{.} \\ \text{Also, acceleration of }y=0, \\ \therefore \sum{{{F}_{y}}=0} \\ \therefore 0={{F}_{N}}-96.78N \\ \Rightarrow {{F}_{N}}=96.78N. \\ \text{It is the force exerted by the normal reaction of the surface on mass}\text{.}$ <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/3617dc0d-e73a-478f-b294-be81731504027984839717002156756.png"/>