A skier of mass kg is pulled up a slope by a motor-driven cable. How much work is required to pull him m up a slope (assumed frictionless) at a constant speed of m/s

Distance covered, s = 8ft - 4ft = 4 ftinitial velocity, u = 34 ft/sthe acrobat is moving upwards and therefore, a = -g = -9.81 m/s^2 = 32.19 ft/s^2applying equation of motion; s = ut -1/2gt^2then,4 = 34t -1/2*32.19t^24 = 34t - 16.095t^2rearranging; -16.095t^2 + 34t - 4 = 0solving the quadratic equation; t = {-34 +/- sqrt [34^2 - 4(-16.)]}/2(-16.095) = 1.06 +/- 0.93 = 0.13 or 1.99the acrobat will take 0.13 seconds to be at 8 ft above the ground. additionally, once at the maximum height, the acrobat will move down and be at the same spot (that is height) after 1.99 seconds moving down.

To compare speeds speeds of wagons we will use second newton's law. it states: f = m * ahere we can see that force depends on the acceleration. if we know acceleration for both wagons we can compare them to make conclusion of speeds of wagons.we have: m1 = 10lbm2 = 20lbf1 = 2*f2also: m2 = 2*m1for bill we have equation: f1 = m1 * a1for tom we have equation: f2 = m2 * a2after inserting identities we have: 2*f2 =m1 * a1f2 = 2*m1 * a2in order to make conclusions about speeds we will solve these equations for a and find their ratio: a1 = 2f2 / m1a2 = f2 / 2m1a1 / a2 = (2f2/m1) / (f2/2m1)a1 / a2 = (2f2*2m1) / (m1*f2)a1 / a2 = (4f2m1) / (m1f2)a1 / a2 = 4this ratio is bigger than 1 and this means that a1 is greater than a2. a1 is 4 times greater than a2. speed of bill's wagon is 4 times greater than speed of tom's wagon.