Amass m = 17 kg is pulled along a horizontal floor with no friction for a distance d =5.8 m. then the mass is pulled up an incline that makes an angle î¸ = 35â° with the horizontal and has a coefficient of kinetic friction î¼k = 0.39. the entire time the massless rope used to pull the block is pulled parallel to the incline at an angle of î¸ = 35â° (thus on the incline it is parallel to the surface) and has a tension t =88 n. 1) what is the work done by tension before the block goes up the incline? (on the horizontal surface.) 418.1 j submit 2) what is the speed of the block right before it begins to travel up the incline? 7.01 m/s submit 3) what is the work done by friction after the block has traveled a distance x = 3 m up the incline? (where x is measured along the incline.) -159.67 j submit 4) what is the work done by gravity after the block has traveled a distance x = 3 m up the incline? (where x is measured along the incline.) -286.67 j submit 5) how far up the incline does the block travel before coming to rest? (measured along the incline.)

Ans : 1) The force pulling the block is 55*cos28Âş; the work done by the force is this times the distance moved: W = 7.2*55*cos28Âş = 350 J 2) The work done by the force appears as kinetic energy of the block W = ½*m*V² = 350J V = âš[2*350/16] = 6.6 m/s 3) The friction force on the block is µ*Fn, where Fn is the normal force of the mass against the plane. Fn = m*g*cos28Âş (the component of weight perpendicular to the plane). Work is the friction force times the distance moved W = 3.3*0.32*16*9.8*cos28Âş = 146 J 4) The force of gravity in the direction of motion up the plane is m*g*sin28Âş, so the work done over a distance of 3.3 m is W = 3.3*16*9.8*sin28Âş = 293 J 5) The kinetic energy at the bottom was computed in 1) above as 350 J. The block will come to rest when the sum of frictional, tension gravitational energy equals that amount. The gravitational energy change moving up the block is m*g*h, where h is the vertical altitude change. Let L be the distance the block moves: h = L*sin28Âş. The friction energy loss is µ*Fn*L; as in 3) Fn = m*g*cos28Âş. The energy supplied by the rope tension is 55*L, so the total energy used moving up the plane is m*g*L*sin28Âş + µ*m*g*cos28Âş*L = 350 + 55*L L = 350/[m*g*(sin28Âş + µ*cos28Âş) - 55] = 5.6 m Alternatively, you can find the acceleration a of the mass from the three forces a = T/m - g*(sin28Âş + 0.32*cos28Âş) = 3.434 - 7.370 = -3.94 m/s² The distance traveled under uniform acceleration -a starting with initial velocity V0 and ending at 0 is s = ½*V0²/a = ½*6.6²/3.94 = 5.5 m

Sorry i don’t know what to do