Question

Calculate the magnitude of the normal force on a 25.2 kg block resting on the floor of an elevator that is accelerating upward at 4.78 m/s 7. a) 367 N b)r! 2 14 Nulingam sit ailarlw .tahn l 'no noianlo004 na zoogrobnu taido ga 0.0 Л c) 612 N d) 913 N .1 8. A 3.00-kg block starts from rest at the top of a 36.0 incline and slides 2.00 m down the incline in 1.75 s. Find the acceleration of the block. a) 1.31 m/s b) 2.11 m/s em 001 d) 11.5 m/s A block of mass 20.2 kg is sliding at an initial velocity of 7.65 m/s in the positive x-direction. The surface has a coefficient of kinetic friction of 0.273. What is the force of kinetic friction (in N) acting on the block? 9. on ni dgiaw i bluow indW .rina® no dl 02, t irlgi sw ingu2ไอ 3ad A a) -16N b) 54N c) -23 N d) -27N " 10. A 6.0-kg bucket of water is raised from a well by a rope. If the upward acceleration of the bucket is 3.8 m/s, find the force exerted by the rope on the bucket. a) 7.60 N b) 12.9N e) 20.2 N d) 81.6N The following relates to questions 11, 12 and 13. A block of mass m-2.50 kg is pushed d-3.00 m along a frictionless horizontal table by a constant applied force of magnitude F 19.0 N directed at an angle 0- 25.0° below the horizontal as shown in the figure below 8 11. Determine the work done by the applied force. a) 0J b) 7.50 c) 51.7 d) 12.7J

Answer 1

Newtons law sigma F_net = m a^vector N - mg = m a^vector N = m(a^vector + g) = 367 N Ans (a) using kinematic S = 1/2 at^2 a = 2s/t^2 = 2 times 2/1.75^2 = 1.31 m/s^2 Ans: - A f = mu N = mu Mg = -0.273 times 20.2 times 9.81 = -54.1 N Ans: - B Newton law sigma F_net = m a^vector T - mg = m a^vector T = m(a^vector + g) = 6 times (3.8 + 9.81) = 81.6 N Ans: - D

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