(1) Mass is released from rest at height above the surface of a table, at the to

Question

(1) Mass is released from rest at height above the surface of a table, at the top of a incline (frictionless and fixed on a table) as shown. Acceleration due to gravity is. (i) Draw the free body diagram for the block (ii) Find an equation for the acceleration of m using g, and (Show your work, if anything missing, you get a zero.) (iii) If = 2.00 kg, = 0.500 m, =30.0 degree find the value of the acceleration. (Show your work) (iv) Find, the magnitude of the normal force. (iv) Find the velocity of the block as it leaves the incline. (Show your work) (v) Find the horizontal velocity component of the block as it leaves the incline. (Show your work) (vi) Find the vertical velocity component of the block as it leaves the incline. (Show your work) Page 3 of 4 (vii) What time interval elapses between when the block leaves the incline and when it hit the floor? Assume (Show your work) (viii) How far from the table will the block hit the floor? (ix) What is the horizontal velocity when the block hit the floor? (x) What is the vertical velocity when the block hit the floor? (xi) Calculate the magnitude of the velocity when it hit the floor???

Explanation / Answer

It will be hard for typing theta using the keyboard . So, i will be writing 'deg' in place of theta . please co-operate and take 'theta' in places of 'deg' .

i)

ii) From the free body diagram , we can see that the force component of 'mg' along the direction i.e., mgsin(deg) is left imbalanced and so , this contributes to the acceleration of the block . We have acceleration = Force/mass . Hence , in this case , we get a = mgsin(deg) / m = gsin(deg) .

im sorry to say this . but a few values are missing from the question , like in part 3 , the variable names are missing and the value of 'H' is not provided . so i will just use their their variable names instead of values .

iii) As long as the block is travelling on the incline , its resultant acceleration would be gsin(deg). (substitute values for answer).

iv) From the free body diagram , we can see that the normal force is balanced by the component of weight perpendicular to the incline i.e., m*g*cos(deg) . So , Noraml Force = mgcos(deg) (substitute values for answer).

v) taking initial velocity (u) = 0 and acceleration = gsin(deg) , the distance travelled along the incline is given by h/sin(deg) . hence using the formula v2=u2+2as , we get v = (2*g*sin(deg)*h/sin(deg))1/2 = (2gh)1/2 . this is the velocity along the direction of the incline . Hence , its horizontal component would be vcos(deg) = (2gh)1/2cos(deg) .

vi) The vertical velocity component would just vsin(deg) = (2gh)1/2sin(deg) .

vii) from the instant it leaves the the incline , the block has to travel a height of 'H' with its vertical component of velocity and experiencing an acceleration 'g' . we have the equation s = ut + (1/2)at2 . substituting values we get , H=(2gh)1/2sin(deg)t + (1/2)gt2 . the required time lapse is the positive solution of this quadratic equation and its big , so comment the values so that i can calculate it clearly for you . :)

viii) let t be the time of travel after leaving the incline as calculated above . the horizontal component of the velocity while leaving the incline was previosuly calculated to be (2gh)1/2cos(deg) . So , the horizontal distance travelled (R) is given by velocity * time = (2gh)1/2cos(deg) * t (note : this t = answer of previous bit)

ix) the horizontal component of velocity doesnot change because there are no forces on the block in the horizontal direction . hence it will remain constant which is previosuly calculated to be (2gh)1/2cos(deg) .

x) we have the equation v = u + at . We have the initail vertical velocity while leaving the incline as (2gh)1/2sin(deg) .

HEnce , final vertical velocity is equal to vf = (2gh)1/2sin(deg) + g*t . (note : again here t is that answer)

xi) the magnitude of velocity is given by v = ( vertical component2+horizontal component2 )1/2

hence substituting , we getv = ( ((2gh)1/2sin(deg))2 + ((2gh)1/2cos(deg))2 ) 1/2 .

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