A 27.9 kg block m_1 is on a horizontal surface, connected to a 6.70 kg block m_2

Question

A 27.9 kg block m_1 is on a horizontal surface, connected to a 6.70 kg block m_2 by a massless string as shown in the Figure. The pulley is mass-leas and frictionless. force of 236.9 N acts on m_1 at an angle of 31.7^degree. The coefficient of kinetic friction between m_1 and the surface is 0.177. Determine the upward acceleration of m_2. What is the velocity of the mass as it loses contact with the spring? 6 278 times 10^-1 7.847 times 10^-1 9.809 times 10^-1 1 226 1 533 1.916 2.395 2.994 A 65.0 kg diver is 1.30 m above the water, falling at speed of 4.60 m/s. Calculate her kinetic energy as she hits the water. (Neglect air friction) 1.05 times 10^3 3.19 times 10^3 9 72 times 10^3 2.18 6 65 2.03 times 10^1 3.16 9.64 2.94 times 10^1 4.59 1.40 times 10^1

Explanation / Answer

15)

when she is 1.3m above the water, she has potential energy due to being above our reference level, and also kinetic

energy due to motion; at this height, her total energy is:

mgh + 1/2 mv^2

where h = 1.3 m and v = 4.6 m/s

when she hits the water, she has no PE and KE in the amount of 1/2mV^2 where V is the speed upon hitting the

water, since the total energy is a constant, we have

mgh + 1/2mv^2 = 1/2mV^2

so her KE hitting the water is:

65(9.81)(1.3) + 1/2(65)(4.6)^2 = 828.1 + 687.7 = 1515.8 J

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