A 200 g object is placed against the free end of a spring...

A 20.0-g object is placed against the free end of a spring (with spring constant k equal to 25.0 N/m) that is compressed 10.0 cm. Once released, the object slides 1.25 m across the tabletop and eventually lands 0.96 m from the edge of the table on the floor, as shown in the figure. Calculate the coefficient of friction between the table and the object. The sliding distance includes the compression of the spring and the tabletop is 1.00 m above the floor level.

Transcript

00:01 Hi, here in this given problem first of all this is the rough horizontal table top over which a compressed spring an object held against this compressed spring having a mass m and then after releasing it it covers a distance over that horizontal table s after which this block, this object is projected in the air with a horizontal velocity.

00:46 Suppose this horizontal velocity is vx and then it falls over the ground.

01:00 Here at this point this horizontal distance x and the vertical height, x, and the so these values are mass of the object 20 .0 gram or we can say 0 .02 kilogram.

01:22 Force constant of the spring 25 .0 newton per meter.

01:31 Compression in the spring delta x 10 .0 centimeter or 0 .10 meter.

01:41 Distance moved over the table 1 .25 meter.

01:47 Horizontal distance over the ground, 0 .96 meter and vertical height covered 1 .00 meter.

01:59 So first of all, using first equation of motion, second equation using second equation of motion for the vertical motion of the object as its vertical initial vertical velocity was zero so h is equal to v y t plus half gt 2 squared plugging in all the known values here for h this is 1 .00 meter is equal to vy t that will become 0 because v y is 0 plus half g 9 .8 into t square which we have to find so this t square will be given by 1 by 4 .9 so the time is calculated to be equal to 0 .45 second time taken.

03:21 By the object to come to the ground.

03:24 So now taking horizontal motion of this object, which is uniform motion.

03:33 So taking horizontal, which is a uniform motion, speed vx will be given by speed equals to distance upon time.

03:52 Distance covered horizontally by it that is given to us.

03:55 Time we have just found.

03:57 So distance was 0 .96 meter, time was 0 .45.

04:03 So it comes out to be equal to speed of the block when it is going to leave the edge of the table, 2 .125 meter per second.

04:17 So here this vx is the final speed.

04:20 Now to find speed of the speed of the speed of the object just after the spring is released.

04:51 So to find it, we use energy conservation, half m v0 square.

05:03 We will consider that speed to be v0 is equal to half k delta x square.

05:13 Cancelling this half, we get an expression...