Mastering Physics: HMwK #5 x C chegg Study Guided se https C Search n.masteringp
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Mastering Physics: HMwK #5 x C chegg Study Guided se https C Search n.masteringphysics.com/myct entProblemID 54771760 emView?assignmen Sessio signed in as alexys barsy Help close Physics 2001 Fall 2015 HMWK#5 Practice Problem 4 2 of 20 next Practice Problem 4.1 SOLUTION In our first example we will consider a very simple SET UP (Figure 1) shows our sketch. We take the +r axis in the direction ofthe horizontal force. "Frictionless" means that the ice doesn't application of Newton's second law. A worker with spikes on supply any force that opposes the motion. Thus, the only horizontal force is the one due to the rope that the worker is pulling. Because the box his shoes pulls with a constant horizontal force of magnitude has no vertical acceleration, we know that the sum of the vertical components of force is zero. However, two vertical forces do act on the box 20 N on a box with mass 40 kg resting on the flat the downward force of the earth's gravitational attraction and the upward normal force exerted by the ice on the box. For completeness, we show these two vertical forces in our sketch; their vector sum must be zero because ay 0 frictionless surface of a frozen lake. What is the acceleration of the box? SOLVE The acceleration is given by Newton's second law. There is only one horizontal component of force, so we have XEFT maar FZ 20 N 20 kg m/s2 0.5 m/s 40 kg 40 kg The force is constant, so the acceleration is also constant. If we are given the initial position and velocity of the box, we can find the position and velocity at any later time from the equations of motion with constant acceleration. REFLECT Two vertical (y) components of force act on the box. its weight w and an upward normal force n exerted on the bottom of the box by the ice-but their vector sum is zero Figure 1 Part A Practice Problem: The box has no vertical acceleration, so the vertical If the box starts from rest and the worker pulls with a force of 30 N,what is the speed of the box after it has been pulled a distance of omponents of the net force sum to zero. Nevertheless, for 0.30 m completeness, we show the vertical forces acting on the box m 40 kg Express your answer in meters per second to two significant figures. In T 20 N m/S Submit My Answers Give UpExplanation / Answer
acc = F/m = 30/40 = 0.75 m/s^2
V^2 = U^2 +2*a*S ( equation of motion for constant acc)
V^2 = 0 + 2*0.75*0.30 = 0.45
=> V = sqrt 0.45 = 0.67 m/s