Two bodies of masses `5kg` and `4kg` are tied to a string as shown If the table and pulley are smooth, then acceleration of `5kg` mass will be .A. `19.5 m//s^(2)`B. `0.55 m//s^(2)`C. `2.72m//s^(2)`D. `5.45m//s^(2)`

Two bodies of masses `5kg` and `4kg` are tied to a string as shown If

1.	Two bodies of masses `5kg` and `4kg` are tied to a string as shown If the table and pulley are smooth, then acceleration of `5kg` mass will be .A. `19.5 m//s^(2)`B. `0.55 m//s^(2)`C. `2.72m//s^(2)`D. `5.45m//s^(2)`
Answer» Correct Answer - D `F =ma, a = (m_(2)g)/(m_(1)+m_(2))`

Discussion

No Comment Found

Related InterviewSolutions

A body rests a rough horizontal plane `A` force is applied to the body directed towards the plane at an angle `theta` with the vertical. The body can be moved along the plane .A. only if `theta` is greater than the angle of frictionB. only if `theta` is lesser than the angle of frictionC. only if `theta` is equal to the angle of frictionD. for all values of `theta`
A particle of mass M rests on a rough inclined plane at an angle q to the horizontal`("sin" theta = (4)/(5))`. It is connected to another mass m as shown in fig. The pulley and string are light. The largest value of m for which equilibrium is possible is M. Find the smallest value of m for which equilibrium is possible
A tall elevator is going up with an acceleration of `a = 4 m//s^(2)`. A 4 kg snake is climbing up the vertical wall of the elevator with an acceleration of a. A 50 g insect is riding on the back of the snake and it is moving up relative to the snake at an acceleration of a. Find the friction force between the elevator wall and the snake. Assume that the snake remains straight.
A block has been placed on an inclined plane . The slope angle of `theta` of the plane is such that the block slides down the plane at a constant speed . The coefficient of kinetic friction is equal to :A. `sin theta`B. `cos theta`C. gD. `tan theta`
A block of mass m slides down an inclined plane of inclination `theta` with uniform speed The coefficient of friction between the block and the plane is `mu`. The contact force between the block and the plane is .A. `mg sin thetasqrt(1+mu^(2))`B. `sqrt((mg sin theta)^(2)+(mumgcostheta)^(2))`C. `mg sin theta`D. `mg`
The force required to just move a body up an inclined plane is double the force the required prevent it from sliding down. If phi is angle of friction and theta is the angle which incline makes with the horizontal then,A. `tantheta=tanphi`B. `tantheta=2tanphi`C. `tantheta=3tanphi`D. `tanphi=2tantheta`
A boy standing on a weighing machine notices his wight as 400 N. When he suddenly jumps upward the weight shown by the machine becomes 600 N. The acceleration with which the boy jumps up is : (Take `g=10m//s^(2)`)A. `5 ms^(-2)`B. `3.4 ms^(-2)`C. `6 ms^(-2)`D. `9.8 ms^(-2)`
A block slides down a frictionless plane inclined at an angle `theta`. For what value of angle `theta` the horizontal component of acceleration of the block is maximum? Find this maximum horizontal acceleration.
A body of mass m starts sliding down an incline of `30^(@)` from rest. The body comes to rest just when it reaches the bottom. If the top half of the plane is perfectly smooth and the lower half is rough, find the force of friction :A. `(mg)/(4)`B. `(mg)/sqrt(3)`C. mgD. `(mg)/sqrt(2)`
In fig., a man true mass M is standing on a weighing machine placed in a cabin. The cabin is joined by a string a body of mass m. Assuming no friction, and negligible mass of cabin and weighing machine, the measured mass of man is (normal force between the man and the machine is proportional to the mass) A. Measured mass of man is `(Mm)/((M+m))`B. Acceleration of man is `(mg)/((M+m))`C. Acceleration of man is `(mg)/((M+m))`D. Measured mass of man is M

Two bodies of masses `5kg` and `4kg` are tied to a string as shown If the table and pulley are smooth, then acceleration of `5kg` mass will be .A. `19.5 m//s^(2)`B. `0.55 m//s^(2)`C. `2.72m//s^(2)`D. `5.45m//s^(2)`

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment