1 we want to slide a 12 kg crate up a 25 m long ramp...

1) We want to slide a 12-kg crate up a 2.5-m-long ramp inclined at 30°. A worker, ignoring friction, calculates that he can do this by giving it an initial speed of 5.0 m>s at the bottom and letting it go. But friction is not negligible; the crate slides only 1.6 m up the ramp, stops, and slides back down. a) Find the magnitude of the friction force acting on the crate, assuming that it is constant. (35 N) b) How fast is the crate moving when it reaches the bottom of the ramp? (2.5 m/s)

Transcript

00:03 Here a crate is given a speed of 5 meter per second at the bottom of the ramp and then it goes up to 1 .6 meter on the ramp and then stops and we have to find the magnitude of friction force on the crate and the speed when it reaches back to the bottom of the ramp so the crate has normal force acting on it component of gravitational force perpendicular to the ramp or the incline which is mg cost 30 degree where m is mass component of gravitational force acting down the incline mg sine 30 degree and the friction force will be acting down the incline when it is going up by 1 .6 meter and friction force is small f now when it is going up we will use work energy theorem work done by all the forces total work will be equal to change in kinetic energy of the great so work done by all the forces will be equal to sum of work done by gravity plus work done by force by friction force plus work done by normal force equal to half m vf square minus v i square now by gravity will be only due to the component of gravitational force parallel to incline because the work done by component of gravitational force perpendicular to incline will be zero because it is perpendicular to displacement so work done by m g cross 30 degree is zero and we will only use work done by m .g.

01:56 Sign 30 degree so displacement is apart and mg sine 30 degree is downward so force and displacements are opposite so work done is negative force minus md g, sine 30 degree times displacement d.

02:15 Similarly, friction force is down the inclined and displacement is up to the incline.

02:20 So work done by friction force is also negative minus fd.

02:24 Worked down by normal force is 0 and is equal to half.

02:28 M v .f is 5 velocity after moving 1 .6 meter which is 0 minus initial velocity square.

02:38 Now we can put the values minus m is 12 kgg g is 9 .8, sine 30 degree and displacement is 1 .6 meter minus friction force is unknown.

02:59 Again displacement is 1 .6 meter equal to half.

03:04 Mass is 12 kg minus initial speed 5 meter per second square...