A body of weight W1 is suspended from the ceiling of a room through a chain of weight W2. The ceiling pulls the chain by a force
Total weight of chain + block system is W1 +W2.
When a constant force is applied to a body, it moves with uniform
From Newton's second law of motion, F= m.a
if force is constant acceleration remains constant.
A particle is moving with a constant speed along a straight line path. A force is not required to
As the particle moves with constant speed in straight line its velocity is constant i.e accleration is zero. From Newton's second law of motion
F=m.a
Force required to keep moving with uniform velocity is zero.
A block is kept on the frictionless inclined surface with angle of inclination . The incline is given an acceleration a to keep the block stationary w.r.t incline plane. Then a is equal to
Taking incline plane as frame of reference, Pseudo force ma act towards right.
component of Pseudo force up the incline is m.a.cos α which is balanced by m.g.sin α. so,
m.a.cos α = m.g.sin α
⇒a= g tan α
A fireman wants to slide down a rope. The rope can bear a tension of (3/4)th of the weight of the man. With what minimum acceleration should the fireman slide down:
According to the question, T=3mg/4
Using equations of motion, i.e. Fnet = m.a
⇒mg-T =ma
⇒mg-3mg/4= ma
⇒a =g/4
A block of mass 10 kg is suspended through two light spring balances as shown in figure
Always remember that spring balance measure Tension in the string in form of kg-wt. As tension in the string is 10 kg-wt.
Reading of Both the scales will read 10 kg
A trolley of mass 8 kg is standing on a frictionless surface inside which an object of mass 2 kg is suspended. A constant force F starts acting on the trolley as a result of which the string stood at an angle of 370 from the vertical (bob at rest relative to trolley) Then:
Using concept of Pseudo force
a= g tanθ
⇒ a = g tan 37 = 10 ×3/4= 7.5 m/s^2
Using Equation of motion
⇒Fnet = 10×2.5 = 25 N
Consider a car moving on a straight road with a speed of 10 ms-1. The distance at which car can be stopped is: [μ=0.5]
Friction force action on the object is f=μ.N= μ.mg
Retardation of the block is a=μ.g = 0.5× 10 =5
Using equation of motion. Stopping distance = u^2/2a=(10 × 10)/( 2* 5)=10 m
A horizontal force of 10 N is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is 0.2. The weight of the block is
Maximum Friction force action on an object is μN.
Here μ=0.2 and Normal reaction force will be equal to applied force i.e 10 N.,
So, friction force is 0.2*10 = 2N
A marble block of mass 2 kg lying on ice when given a velocity of 6 ms-1 is stopped by friction in 10 s. Then the coefficient of friction is
Using equation of motion
v= u + at
⇒ 0= 6 + a ×10
⇒ a= -0.6 m/sec^2
Maximum Friction force is μN = μ mg so, a= μ.g
⇒ -0.6 = - μ. 10
⇒μ = 0.06