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Car mass is 1200 kg moving velocity is 108 km/hr. On applying brakes velocity reduced to 36 km/hr. Find change in momentum?
(posted by Aditi)
u=108 x 5/18=30m/s
Change in momentum = m(v-u) = 1200 x (30-10)=1200 x 20= 24000 kg m/s
“The friction coefficient between a road and the tyre of a vehicle is 4/3.find the maximum incline the road may have so that once hard brakes are applied and the wheel starts skidding,the vehicle going down at a speed of 10m/s is stopped within 5m” Raju Rabha asked
Two blocks A and B of mass m1 and m2 respectively are kept in contact on a frictionless table.The experimenter pushes the block A from behind so that the blocks accelerate. If the block A exerts a force F on the block B,what is the force exerted by the experimenter on A.
I just want to know how to easily solve problems related to kinematics and friction. I don’t know why i always face problem in these two topics a lot……
Please help and tell some tricks to solve these both…….
“Can you suggest a way to tackle problems on constraint relations easily??” – Aritra posted this question.
Answer: The common way to solve problems involving constraints is to replace the constraints with their reaction forces. As the question is not specific, I cannot tell much now. If you have further doubts regarding this issue, please post as comments to this post.
“What is resultant force?”
When a number of frces are acting simultaneously on a body, it will be producing an effect – may be keeping the body in equilibrium, or giving it an acceleration in a particular direction.
Resultant of a number of forces is defined as that single force which can replace all other forces to produce the same effect.
If the number of forces acting on the body keep it at rest or maintain uniform motion,it is said to be in equilibrium; then the resultant is zero (since acceleration is zero).