00:01
Here in this question we assume that the skater and incline as a system and draw a free body diagram of the system by identifying the forces acting on it.
00:15
So let's draw a free body diagram of the skater on the incline.
00:22
So the given figure shows the free body diagram of the skater on the incline.
00:32
Now in the given figure, small m represents the mass of the skater, theta is the angle of inclination, mg sine theta, mg post theta are the components of the weight of the skater, fk be the kinetic frictional force and capital n be the normal friction force.
00:58
So now here we assume that the acceleration of the skater on the incline b, a1.
01:26
Now we are going to calculate the net force on the system by applying newton's second law along x direction.
01:34
So, summation fx net equals to m .a .1.
01:47
Now, let's resolve this.
01:49
So we get mg sine theta plus fk equals to m multiply by a1.
02:00
Now we name this equation as equation first or one.
02:05
Now, similarly we calculate the net force on the system by applying newton's second row along the y direction.
02:16
Therefore, it means summation fy net equals to m .a1.
02:29
Now let's again resolve this.
02:32
So we get capital n minus.
02:38
M g pose theta equal to zero now since there is no motion along vertical direction so acceleration a1 is equal to zero so from above we get n is equal to m g pose theta now we name this equation as equation two or second now the expression for the kinetic frictional force is given by fk equal to mu k multiplied by capital n now here mu k is the coefficient of kinetic friction so from equation to we get f k equals to mu k multiply by m g cos theta now we substanti substitute this equation in equation first and calculate the acceleration of the skater.
04:22
Therefore, we get mg sine theta minus mu k m g post theta equal to m multiply by a1.
04:42
So from above we get a1 equals to g multiply by sine theta minus mu k cost theta now here we substitute 9 .80 meter per second square for g 0 .18 for mu k and 28 degree for theta therefore a1 equals to 9 .8 multiplied by sine 28 degree minus 0 .18 multiplied by cos 28.
05:21
So we get a1 equal to 3 .04 meter per second square.
05:32
Hence the acceleration of the skater on the incline is 3 .04 meter per second square.
05:40
Now we are going to calculate the final speed of the skater on the incline by using kinematics equation.
06:01
So we know that v square minus v0 equal to 2k1 multiply by d.
06:12
Now here v is the final speed v note is the initial speed and d is the length of the incline so we rearrange this for v therefore v is equal to square root of v0 square plus 2a1 multiply by d now here we substitute 5 meter per second for v note 3 .0 4 meter per second square for a 1 and 110 meter for d therefore v equals to square root of 5 whole square plus 2 multiply by 3 .04 multiply by 110...