The gigantic globe (mass 500kg) from atop the Daily planet Headquarters has fall

Question

The gigantic globe (mass 500kg) from atop the Daily planet Headquarters has fallen down to the ground, and superman has arrived to clean up for us. The globe is still a sphere of radius r= 3.5m and is lying touching both the ground and a 50 degree incline, superman stretches to warm up and then applies a force P which is pointed at an angle directly towards the center of the globe.

a. If his hands are at a height h=1.54m, what is the minimum force P that he must apply in order to start the mass sliding up the incline. Assume the coefficient of static friction between the globe and the incline is us=.65 hint: the globe will start up the incline when the normal force is exerted on the globe by the horizontal surface becomes zero and the static frictional force is maximized.

b. If, after starting the mass from rest, superman's pushing force P linearly increases its magnitude to 1.5 times its part (a) value over a distance of 1.00m, stays at that value for the next 1.5m, and then decreases back to zero over the last .5m of the incline, how fast is the 500kg globe going at that point (total distance traveled: s= 3.00m)? Assume the direction of P, i.e. remains constant during the motion.

c. How fast would the globe be going if there still was a kinetic frictional force? Assume uk= 0.54

Explanation / Answer

a)the force is Fx = w * sinA - f_s = w * sinA - u_s * n Fy = n + (-w * cosA) = 0 or n = w * cosA or Fx = w * sinA - u_s * w * cosA = w * (sinA - u_s * cosA) where w = mg,m = 500 kg and g = 9.8 m/s^2,A = 50o and u_s = 0.65 b)the force apllied is Fx' = 1.5 * Fx the work done by the force is W = Fx' * S where S = 3.00 m let the speed of the globe at that point be v we know that W = (1/2)mv^2 or Fx' * S = (1/2)mv^2 or v = (2Fx' * S/m)^1/2 c)when there was still a kinetic frictional force then Fx = w * (sinA - u_k * cosA) or Fx' = 1.5 * w * (sinA - u_k * cosA) where u_k = 0.54 or v' = (2Fx' * S/m)^1/2

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