m The object of mass m = 3 kg is at the bottom of the incline, which is at the angle e = 55° with respect to horizontal. The object has initial speed of vo = 5 m/s and begins slide up the incline. There is a friction between the object and the incline. The coefficient of kinetic friction is uk = 0.25. The object moves up the incline and eventually stops under the influence of friction and its own weight. Find: (a) how high up the incline (in m, relative to its original position) the object slides before stopping (careful here - your answer is the height, i.e. the vertical distance between object's final position and the ground); (b) work in J) done by the kinetic friction when the object moves up the incline (careful with the sign); (c) work in J) done by the normal force when the object moves up the incline; (d) work in J) done by the gravity when the object moves up the incline (careful with the sign); (e) when the object reaches the highest point on the incline it begins to slide back down, find object's speed (in m/s) at the bottom of the incline.

(a) First, we need to find the distance the object slides up the incline before stopping. We can use the principle of conservation of energy for this. The initial kinetic energy of the object is (1/2)mv^2, and this will be converted into potential energy (mgh) and work done against friction (W_friction = ukmgdcos(90-e)) as the object moves up the incline.

So, (1/2)mv^2 = mgh + ukmgdcos(90-e)

Substituting the given values, we get:

(1/2)3(5^2) = 39.8h + 0.2539.8dcos(35)

Solving for h (height), we get h = 1.28 m.

(b) The work done by kinetic friction is given by W_friction = ukmgdcos(90-e). Substituting the given values, we get:

W_friction = 0.2539.8dcos(35)

The distance d the object slides up the incline can be found using the relation d = h/sin(e). Substituting h = 1.28 m and e = 55°, we get d = 1.56 m.

Substituting d = 1.56 m in the equation for W_friction, we get W_friction = -11.4 J. The negative sign indicates that the work is done against the motion of the object.

(c) The work done by the normal force is zero, because the normal force is perpendicular to the direction of motion.

(d) The work done by gravity is given by W_gravity = -mgh. Substituting the given values, we get:

W_gravity = -39.81.28 = -37.44 J. The negative sign indicates that the work is done against the motion of the object.

(e) When the object slides back down the incline, it gains kinetic energy equal to the potential energy it had at the top of the incline. So, its speed at the bottom of the incline can be found using the equation (1/2)mv^2 = mgh. Solving for v, we get:

v = sqrt(29.81.28) = 5.04 m/s.