A block of mass m = 7.90 kg is released from rest at a height of H = 3.00 m on a

Question

A block of mass m = 7.90 kg is released from rest at a height of H = 3.00 m on a curved frictionless ramp. At the foot of the ramp is a spring whose spring constant is k = 221.0 N/m. What is the maximum compression of the spring, x? A 1102 kg block of granite is pulled up an incline that has an angle of inclination of 34.0 degree with a constant speed of 2.23 m/s by a steam winch (see Figure). The coefficient of kinetic friction between the block and the incline is 0.260. How much power must be supplied by the winch? The string in the Figure is L = 108.0 cm long and the distance d to the fixed peg P is 89.6 cm. When the ball is released from rest in the position shown, it will swing along the dashed arc. How fast will it be going when it reaches the lowest point in its swing? How fast will it be going when it reaches the highest point in its swing (after the string encounters the peg)? In a rope climbing competition, a 68.2 kg athlete climbs a vertical distance of 5.16 m in 9.76 s. What minimum power output was used to accomplish this feat?

Explanation / Answer

2. 5 joules. W = Fd and between 0 and 1 m you have a 2 N force, that accounts for 2 joules. Between 1 and 2 m you have an average force of 3 N, that makes another 3 joules for a total of 5. Between 2 and 3 m there is no force.

3. mgh = mv2/2 so v = (2gh)1/2 = (2 * 9.81 m/s2 * 1.08 m)1/2 = 4.60 m/s

4. We can see that r = 1.08 - .896 = 0.184 m, so that potential energy difference between the start and the high point is not mgh but mg(1.08 - 2 * 0.184); so v = (2g * 0.712 m)1/2 = 3.74 m/s

5. Change in potential energy is (68.2 kg)(5.16 m)(9.81 m/s2) = 3452.26 J so power is (3452.26 J)/9.76 s = 353.7 w

6. Every second the block gains (2.23 m/s)(sin 34) = 1.247 m of height so its potential energy is increasing by (1102 kg)(1.247 m)(9.81 m/s2) = 13480.8 J. Additionally, there is a frictional force of (1102 kg)(9.81 m/s2)(cos 34)(0.26) = 2330.2 N which must be overcome by the winch; every second this force is applied over 2.23 m so another 5196.4 J is expended. The total energy output in one second is 18677.2 J so the power is 18677.2 w.

7. kx2/2 = mgh so x = (2mgh/k)1/2 = ((2 * 7.9 kg)(9.81 m/s2)(3.00 m)/(221 N/m))1/2 = 1.45 m

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