A particle moves from position 3i+2j-6k to 14i+13j+9k due to a uniform force of (4i +j + 3k) N. If the displacement is in meters, then will work be done?
Work done =force×displacementGiven data,force f=4i+j+3k NNow displacement,d=(3i+2j-6k)~(14i+13j+9k)d=(14–3)i+(13–2)j+(9-[-6])kd=11i+11j+15kNow,W=f×dW=(4i+j+3k)×(11i+11j+15k)W=(11×4)+(1×11)+(3×15) °.° [i×i=1]W=44+11+45W=100 joules
A 50-kg box is pushed along a horizontal surface. The coefficient of kinetic friction is 0.35. What horizontal force would accelerate it at 1.2 m/s^2?
The situation involves taking into account two main contributing forces. These are the Friction Forces and the Accelerating forces.1. If the box has to be moved with constant velocity then only the frictional forces will be existing. ( 50. 9.8. 0.35) this assumes that the frictional forces remain constant and that the force is horizontal. If the force was at a down or up angle pushing or at an up or down angle pulling then the problem would add other components due to the up angle or down angle of the pushing force.2. When accelerating in addition to the friction forces one also needs to add an accelerating force which is M.a or (50. 1,2)Again it would add a little more effort if one considers the pushing or the pulling forces, not only when horizontally applied but at an angle. This would give an idea of what to select in actual real life when one is faced with this dilemma.
Two blocks are placed in contact on a smooth horizontal surface. Their masses are 2 kg and 3 kg. A 100 N force pushes on the 2 kg block. What is the magnitude of the force that the 2 kg block exerts on the 3 kg block?
This is a basic problem based on Newton's Laws of Motion. Refer the below given pics for the detailed solution. The same problem can be solved by proceeding in slight different manner also i.e. by applying Newton's second law separately on both the blocks; thus getting two equations with two unknowns (acceleration and mutual force). One more thing to note here that while solving the problem the external force of 100N has been assumed to act on the 2kg block in Horizontal direction as it has not been clearly specified in the statement of the problem.Regards!
If two blocks having mass (m1) and (m2) are pushed with a force F on a frictionless surface and begin moving with a uniform acceleration 'a', what would be the force between the two blocks?
The question can be solved by simple application of Newton's Second Lawof Motion.Using Newton's Second Law,it is established that[math]F_{ext}=m_{com}\,a[/math],where [math]F_{ext}[/math] is net external force acting on the system, [math] a[/math] is the acceleration produced and [math]m_{com}[/math] is mass (of center of mass of) the system(which is basically total mass of the system).Now consider both [math]m_{1}[/math] and [math]m_2[/math] as a system, so net external force acting on this system is [math]F[/math] and[math]m_{com}=m_1+m_2[/math].[math]\Rightarrow F=(m_1+m_2)a[/math][math]\Rightarrow a=\dfrac{F}{m_1+m_2}[/math].Now to find normal force between two blocks we have to consider individual blocks as a system because if you consider both of them as a single system, the normal force becomes an internal force and Newtons's Law just talks about external force.Also note that from Newton's Third Law, the normal force will act on both blocks and will be of same magnitude but in opposite direction.So lets assume that a normal force [math]F_n[/math] acts on the system.Clearly this normal force will drive [math]m_2[/math] but will oppose motion of [math]m_1[/math] (Why?).So if you consider [math]m_2[/math] as your system, net external force is [math]F_N[/math] ,mass of (center of mass of) the system is [math]m_2[/math] and acceleration is [math]a[/math].[math]\Rightarrow F_n=m_2a[/math][math]\Rightarrow \boxed{F_n=\dfrac{m_2F}{m_1+m_2}} [/math].And if you consider [math]m_1[/math] as your system,net external force is [math]F-F_N[/math] , mass of( center of mass of) the system is [math]m_1[/math] and acceleration is [math]a[/math].[math]\Rightarrow F-F_n=m_1a[/math][math]\Rightarrow F_n=F-m_1a[/math][math]\Rightarrow \boxed{F_N=\dfrac{m_2F}{m_1+m_2}}[/math]Hence required answer is [math]\boxed{\dfrac{m_2F}{m_1+m_2}}[/math].Few questions for you:-Is the given condition [math]m_1>m_2[/math] relevant? Or the given question can be solved if this condition is not given.Why is acceleration of both the masses combined is same as acceleration of individual mass?
A force of 100 N is acted on a block with a of mass 50 kg and produces a constant velocity of 5 m/min on a rough surface. Can you find the force of friction?
It is a very interesting real fact that Force has a value if and only if the force applied on any body is moving with an acceleration i.e the velocity should get increasing. If not Force is simply Zero.Force = Mass X AccelerationIf Velocity is constant or not changing, then Acceleration is absolute 0Resulting Force = Zero.In this case, when the Applied force 100 N was acted on the box, initially the Net force had a value until the box accelerated to 5 m/min (motion) from 0 m/min (rest). Once the box attained the velocity of 5 m/min and became constant, the Net force became Zero.So, Net Force = Applied Force - Frictional ForceAs per above Explanation Net Force becomes 0 as the velocity becomes constant.0 = Applied Force - Frictional ForceApplied Force = Frictional Force = 100 N (Answer)
How do I calculate the average braking force on a car of mass 800 kg initially moving at 8.9 m/s?
T To answer this question we require the time .Lets assume its one second.Mass = 800 kgInitial velocity = 8.9 m/sFinal velocity = 0 m/sTime = 1 sBRAKING FORCE = Mass × AccelarationAcc. = v-u ÷t= -8.9Thus , force = 7120
A car of mass 1500 kg accelerates from rest at 5 m/s2 for 5 s on a straight road. What is the minimum power needed for the engine to achieve the task?
what Sol Infinus, did is correct. But to make sure, here is a longer way of finding the same colutionpower = work done / timeWork done = force x distanceF = ma5 x 1500 = 7500to calculate distance we can use SUVATv = u +at0 + 5 x 5 = 25v^2 = u^2 + 2as25^2 = 0 + 10s625 / 10 = 62.5again. Work done = force x distance7500 x 62.5 = 468750power = work done / time468750 / 5 = 9375wor 93.75KWthat is average power187.5KW for final power is correct for power needed to reach a velocity of 25that is: power = force x velocity 7500 x 25 = 187.5KW
If a car drives with a speed of 80 km/h and suddenly the driver reacts and pushes on the brakes with a reaction time of 0.6 seconds and the acceleration is -6.2 m/s^2 during the braking, how do I calculate the braking distance and the shutdown distance?
Using kinematics equation, v^2=v0^2+2gx, where v is the final velocity, v0 is the initial velocity, g is the acceleration which is negative in out case and x, which is the distance to get the car from 80km/h to 0km/h with the given deacceleration. Thus v is 0. Solving for x will give the distance required for the car to come to an complete stop. Thats for the first part and for the second part, we can get a distance the car travel within the reaction time,t by multipling the reaction time with car vocity. This is the distance the car covers before the brake is initiated. Adding this distance and the distance from first part will give you the total distance car requires to come to a full stop, that's with the reaction time included. Make sure to convert all units to m/s.