For this problem on the topic of energy conservation, we're told that a 0.5 kg mass is propelled up an incline by the use of a spring with spring constant 500 newtons per meter. The spring is initially compressed, but that he centimeters from its equilibrium position and sends the mass from rest firstly across a horizontal surface and then onto the plane. The plane has a length of four m an incline of 30° to the horizontal. The coefficient of kinetic friction between the mass and the surface is 0.35. And When compressed this, when the spring is compressed, the masses 1.5 m from the bottom of the plane. We want to find the speed of the mass as it reaches the bottom of the plane. The speed of the mass as it reached the top of the plane. And the total work by friction from the beginning to the end of the masses motion. Now the elastic potential energy is you spring. Yeah. And this is equal to half K X squared where K is the spring constant? And X is the compression. The mass loses energy which is the work done by friction, which is minus the frictional force, F F times the distance that the mass traveled. D. And this is equal to minus UK times M G. D. Due to friction. And therefore the kinetic energy at the bottom is given by K B. And K B is equal to a half M. The B squared which is equal to half K X squared minus the work done due to friction, Mieux, que times am times G times D. And so from here we can rearrange to find the speed of the block at the bottom of the plane. VB to be the square root of K X squared over two K X squared over em rather minus two UK jean times deep. And so since all these values are known if we substitute them in, we get this to be the square root of 500 newtons per meter times the compression of the spring, 0.3 m squared divided by The mass of the block, 0.5 Kg -2 times. The coefficient of kinetic friction is 0.35 Times the acceleration due to gravity 9.81 m/km2 Times a distance of 1.5 m which gives us the speed of the block at the bottom To be eight 93 meters per second. So that's the block speed just before it starts moving up the inclined plane. Now, for part B we have that to reach the top of the incline. The gravitational potential energy must also be considered. And the change of gravitational potential energy, delta U. G. Is equal to the potential energy at the top minus the potential energy at the bottom. Now, since the plane has lent l the an incline angle theater, the change in gravitational potential energy is MGl scientist to. And the kinetic energy at the top can then be calculated by subtracting the gravitational potential energy and work due to friction from the kinetic energy at the bottom. So Kay top is equal to okay at the bottom minus UK MG L. Sign theater minus mgl sine theta. And so by rearranging this equation, since we know half um the top squared is equal to K B minus UK mgl signed data minus MG al sine theta. We can find the speed of the block at the top of the incline, the top to be the square root of two over. Em into the kinetic energy at the bottom minus M G L. Into UK. Call sign data plus sign data. And so if we substitute the values into this equation, we get this to be the square root of two over 0.5 Kg. And the kinetic energy at the bottom. We can get from VB, which we calculated in part A and it's a half M V B squared which is 19 0.92 jules, yeah -0.5 Kg In 29.81 m/km2 times for meters multiplied by 0.35 Times The co sign of 30° plus the sign of 30 degrees. And so if we calculate this, we get the speed at the block at the top To be 4.08 m per second. And lastly, we want to find the total work done by friction from the beginning to the end of the motion of the mass. Now the total work done due to friction is equal to minus the frictional force ff in two D plus L, which is minus UK m g d minus UK mg. Call sign of theater times L. And so if we substitute our values into this equation, this is minus 0.35 times zero 0.5 kg Times 9.81 m/km2 Into 1.5 m less, four m times the call sign of 30 degrees. And so this gives us the work done due to friction for the blocks, entire motion to B -8 0.52 jules.