Free Study Material

NCERT Solutions

NCERT Books

Revision Notes

Practice Problems

This is one of our many quizzes on Net force. Click on this to access more such competitive quizzes on Mechanics.

Net force Quiz 2

physics

India’s longest train journey spans from Dibrugarh in Assam to Kanyakumari in Tamil Nadu. You decide to go on this journey that covers a distance of about $4000\,km$ in five days, travelling through nine states. Towards the end of your journey, you decide to get a cup of coffee from one of the halt stations to freshen up before concluding your trip. You bring a $250\,g$ cup of coffee and place it on a tray inside the train compartment.  If the train starts to exit the station at the rate of $4.5\,ms^{-2}$, determine the net force acting on the cup if the coefficient of static friction between the cup and the tray is $0.45$.  

First find the force produced by the acceleration of the train and then find the magnitude of the frictional force. Use this to arrive at the net force acting on the cup. 

 We are given that the mass of the coffee cup is $m = 250\,g = 0.25\,kg$,  $ \\ $ And the acceleration of the cup is $a = 1.5\,ms^{-2}$ $ \\ $ The force on the cup due to this acceleration will be: $F = ma \\ \Rightarrow F = 0.25 \times 4.5 \\ \Rightarrow F = 1.125N \\ $ Given that $\mu_s = 0.45$, the frictional force acting on this cup will be: $ \\ F_{r} = \mu mg \\ \Rightarrow F_{r} = 0.45 \times 0.25 \times 9.8 \\ \Rightarrow F_{r} = 1.1025N \\ $ Therefore, the net force acting on the cup, given that frictional force always opposes motion, will be: $ \\ F_{net} = F – F_{r} \\ \Rightarrow F_{net} = 1.125 – 1.1025 \\ \therefore F_{net} = 0.0225\,N$

Suppose you and your friend decide to go to the arcade and spend an afternoon together playing at the bowling alley. You pick up a $5\,kg$ bowling ball and hurl it across the alley by rolling it towards the pins with a force of $7.5\,N$. If the coefficient of rolling friction of the ball is $0.08$ determine the net acceleration of the bowling ball as it approaches the pins.  

Determine the initial acceleration imparted to the ball due to the force exerted on it, and then find the deceleration due to friction to arrive at the appropriate net acceleration.

  Given that you exert a force $F = 7.5\,N$ on the bowling ball of mass $m = 5\,kg$, $ \\ $ The acceleration imparted to the ball will be: $ \\ a = \dfrac{F}{m} \\ \Rightarrow a = \dfrac{7.5}{5} \\ \Rightarrow a = 1.5\,ms^{-2} \\ $ Given that $\mu = 0.08$ $ \\ $ Let $F_{r} = \mu mg$ be the frictional force acting on the ball as it rolls down the aisle. $ \\ $ The net force acting on the ball will be: $ \\ F_{net} = F-F_{r} \\ \Rightarrow ma_{net} = ma - \mu mg \\ $ Therefore, the net acceleration will be: $ \\ a_{net} = a - \mu g \\ \Rightarrow a_{net} = 1.5 – (0.08 \times 9.8) \\ \therefore a_{net} = 0.716 \approx 0.72\,ms^{-2}$

 You decide to go to a shooting range to do some target practice for the upcoming pheasant season. The targets at the range are made of self-healing rubber. You aim and pull the trigger when the slug exits the gun barrel and strikes the rubber target. If the $0.05\,kg$ slug decelerates at the rate of $2.5\,ms^{-2}$ as it travels through the rubber, determine the net force experienced by the slug. $\text{(Take the coefficient of friction between the bullet surface and rubber to be }\mu = 0.8)$. 

 First calculate the retarding force, and then the frictional force to eventually arrive at the appropriate net force. Note that the initial two forces act along the same direction.  

Given that a slug of mass $m = 0.05\,kg$ decelerates at the rate of $a = -2.5\,ms^{-2}$.  $ \\ $ The retarding force on the slug due to this deceleration will be: $ \\ F = ma \\ \Rightarrow F = 0.05 \times (-2.5) \\ \Rightarrow F = - 0.125\,N \\ $ Now, $\mu = 0.8$, therefore the frictional force acting on the slug will be: $ \\ F_{r} = \mu mg \\ \Rightarrow F_{r} = 0.8 \times 0.05 \times 9.8 \\ \Rightarrow F_{r} = 0.392N \\ $ Therefore, the net force experienced by the slug will be: $ \\ F_{net} = F – F_{r} \\ \Rightarrow F_{net} = -0.125-0.392 \\ \therefore F_{net} = -0.517 \approx -0.52\,N$

 Located in Alberta, Canada, the Abraham Lake is known for its saturated blue colour and the frozen methane bubble trains that are formed underneath the lake’s surface as the ice sets in. Suppose that an ice skater of mass $60\,kg$ is skating over this lake. If the skater, initially at rest, reaches a velocity of $10\,ms^{-1}$ in $4\,s$ and the coefficient friction between the blades of the skate and the ice is $0.1$, determine the net force acting on the skater.  

Begin by determining the acceleration of the skater for the rate at which she arrives at the given velocity and then find the net force while accounting for friction.  

Given that the skater’s initial velocity is $u = 0\,ms^{-1}$ and final velocity is $v = 10\,ms^{-1}$ and $t = 4\,s$, the acceleration of the skater can be found to be: $ \\ a = \dfrac{v-u}{t} \\ \Rightarrow a = \dfrac{10-0}{4} \\ \Rightarrow a = 2.5\,ms^{-2} \\ $ Given that the mass of the skater is $m = 60\,kg$, the force she experiences due to the above acceleration will be: $ \\ F = ma \\ \Rightarrow F = 60 \times 2.5 \\ \Rightarrow F = 150\,N \\ $ Now, if $\mu = 0.1$ then the frictional force acting on the blades of the skater’s shoes will be: $ \\ F_{r} = \mu mg \\ \Rightarrow F_{r} = 0.1 \times 60 \times 9.8 \\ \Rightarrow F_{r} = 58.8N \\ $ Therefore, the net force acting on the skater will be: $ \\ F_{net} = F-F_{r} \\ \Rightarrow F_{net} = 150-58.8 \\ \therefore F_{net} = 91.2\,N$

A canned foods factory uses a conveyor belt system to transport the goods from the manufacturing level at the bottom to the packaging level at the top of the building as shown in the figure. If a single can weighs $500\,g$ and accelerates at the rate of $3.5\,ms^{-2}$, determine the absolute coefficient of friction between the conveyer belt and the can if the net force acting on the can is zero. $ \\ $  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/4420151e-ed9f-4d07-af9b-8e1f908a45e08232313293217182301.png"/>

Sketch out a free-body diagram to determine the influencing forces and their directions. Use this to arrive at an appropriate equation for the net force, which you can equate to zero and solve.  

 Given that the mass of a can is $m = 500\,g = 0.5\,kg$ and accelerates at $a = 3.5\,ms^{-2}$. $ \\ $ We begin by sketching out a free-body diagram and resolving the gravitational force into its respective components as shown in the figure. $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/78debcc8-dec7-44d1-ae7c-c4e0768e37ca1796828926651502582.png"/> $ \\ $  From the diagram, we see that the force on the can due to the conveyor belt will be: $ \\ F = ma \\ \Rightarrow F = 0.5 \times 3.5 \\ \Rightarrow F = 1.75\,N \\ $ Let the coefficient of friction between the conveyer belt and the can be $\mu$. $ \\ $ We also see that frictional force on the can will be: $ \\ F_{r} = \mu N \\ \Rightarrow F_{r} = \mu mgcos 30^{\circ} \\ \Rightarrow F_{r} = \mu \times 0.5 \times 9.8 \times 0.866 \\ \Rightarrow F_{r} = \mu 4.24N \\ $ Given that the net force acting on the can as it moves on the conveyor belt is zero, i.e., $ \\ F – F_{r} – mgsin30^{\circ} = 0 \\ \Rightarrow 1.75 – \mu 4.24 – (0.5 \times 9.8 \times 0.5) = 0 \\ \Rightarrow \mu 4.24 = 1.75 – 2.45 \\ \therefore |\mu| = \dfrac{|-0.7|}{4.24}= 0.165$