A child is playing with a toy car of mass 025 kg along a...

A child is playing with a toy car of mass 0.25 kg along a smooth inclined plane which makes an angle 30 degree with the horizontal. ( 30 points) a. The car stays at rest along the inclined plane. Find the parallel and perpendicular components of the weight of the car. b. At what angle, the parallel and perpendicular components will be equal. c. At the end of 5.00s find the final velocity of the car at the bottom. d. If the car takes 6.5 s to reach to the bottom from the top, find the length of the incline. e. Now, the child spills dirt on the incline, the coefficient of friction between the car wheels and the inclined surface is 0.5. Find the acceleration f. Find the new velocity at the end of 5.00 s g. Find the time the car takes to reach the bottom. h. What force must be applied to move the car upward with a uniform velocity? i. What is the physics involved when the angle of the incline is increased to 90 degrees? j. Draw the free body diagram involving all the forces in the entire problem.

Transcript

00:01 Hello, the question is taken from physics and the question is a child is playing with a toy car of mass 0 .25 a long way smooth inclined plane which makes an angle of 30 degree with the horizontal.

00:12 Okay, so this is smooth inclined plane and it is making it an angle 30 degree with the horizontal.

00:21 Okay.

00:23 And mass is 0 .25kj and the current.

00:31 Car stays at rest along the inclined plane find the parallel and perpendicular component of the weight of the car so weight is this so one component is this is 0 .25 into gg is acceleration due to gravity this is perpendicular component and parallel component is in this direction so one component is parallel component so weight parallel so that will be 0 .25 g into course 30 degree so that will be 0 .25 into 9 .8 into square root 3 by 2 so that will give this value 0 .25 into 9 .8 into square 3 by 2 so that will be second 0 .25 into 9 .8 into square 2 by 2 so that will be 2 .12 newton and in second 0 .25 into 9 .8 into square 3 by 2 so that will be 2 .1 2 newton and in second case perpendicular component of weight will be 0 .25 into 9 .8 into sine 30 degree so that will be 0 .25 into 4 .9 so that will give us the value of weight is 0 .25 into 4 .9 so that will be 0 .05 newton which is the required value of the weight and in second part what they are asking at what angle the parallel and perpendicular component are equal so that is w parallel pose of w parallel at what angle is equal to w perpendicular which is equal to m g course theta is equal to m g sign of theta this is possible only when theta is equal to 45 degree and third part is at an angle of five second what is the final velocity of the car at bottom okay, so initial velocity is equal to i'm taking that initially it is at rest so we is equal to is g sine of theta into t is equal to 5 second so we will be g is 9 .8 sign into 5 so that will be 4 .9 into 5 which is 24 .5 meter per second and 4 .5 parties if the car take 6 .5 second to reach the bottom of the top find the length of the incline so l is equal to v into t so t that is given to us which is 6 .5 second so that will be 9 .8 sign 30 degree into 6 .5 square so that will be so let me value at this value of land velocity into time which is 9 .8 sine 30 degree multiply 6 .5 square so that will be 207 meter and fifth part is now the child spills the dirt on the incline the coefficient of friction between car and the inclined surface is 0 .5 find the acceleration at this point acceleration is equal to g sine theta minus mu g course of theta that is a friction is friction forces mu into m g course of theta where this is equal to n okay so acceleration become m g course of theta so a will be 9 .8 into sine theta is sine 30 degree 0 .5 course 30 degree so let us evaluate the value of acceleration which is course sorry 0 .5 minus 0 .5 into square 3 by 2 into 9 .8 so that value is 0 .656 meter per second square and 6 part is find the new velocity at the end of 5 second b is equal to u plus 80 u initial velocity what was that value 24 .5 .5.

05:46 Meter per second plus that was the initial velocity in the absence of any friction force i think i should take initial velocity is equal to zero because the friction is present from the center initial value so acceleration is your point at the end of five second into five so that will be 3 .28 meter per second seventh part is find the time take time the car takes to each the bottom t is equal to v by a velocity at the end of a sorry when it reach at the bottom find the time the car takes to reach the bottom length we know no need of this v is equal to u plus a t v no we don't know the value of we at that time so as is equal to u t plus half 80 square we know the initial velocity s is equal to u t plus half 80 square but we don't know the value we don't know the value of you is equal to zero initially i'm taking it at rest so t will be square two into l over a we know the length of the inclined plane so a l is equal to 207 2 into 207 acceleration is 0 .656 so t will be square root 2 into 207 divided by 0 .656.

08:12 So that will be 25 .12 second...

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