Homework Set 3: Solutions Due: Tuesday, September 14, 2010 ...

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(E4) 6 pointsA 3.0-kg block being pulled across a table by a horizontal force of 80 N also experiences africtional force of 5 N. What is the acceleration of the block?The net force acting on the block is ∑ −→ F = 80 N−5 N = 75 N in the direction of the pull. Therefore,using Newton’s second law, the acceleration of the block is∑−→−→ F a =m = 75 N = 25 m/s23.0 kg(E8) 10 pointsin the direction of the pull.A 4-kg block is acted upon by three horizontal forces as shown in the diagram.a. What is the net horizontal force acting on the block?The net horizontal force acting on the block is ∑ −→ F = 25 N+5 N−10 N = 20 N to the right.b. What is the horizontal acceleration of the block?The horizontal acceleration of the block isto the right.(E12) 10 points Jennifer has a weight of 110 lb.−→ a =∑−→ Fm = 20 N4 kg = 5 m/s2a. What is her weight in newtons? (1 lb = 4.45 N)In newtons, her weight isb. What is her mass in kilogram?In kilograms, her mass is( ) 4.45 N110 lb = 110 lb = 490 N.1 lbm = W g = 490 N2= 50 kg.9.8 m/s(E18) 10 points An upward force of 18 N is applied via a string to lift a ball with a mass of 1.5 kg.a. What is the net force acting upon the ball?The net force acting on the ball is∑F = T −W = T −mg = 18 N−(1.5 kg)(9.8 m/s 2 ) = 3.3 Nin the upwards direction.b. What is the acceleration of the ball?The acceleration of the ball isin the upwards direction.∑−→−→ F a =m = 3.3 N = 2.2 m/s21.5 kg
(SP6) 15 points A 60-kg man is in an elevator that is accelerating downward at the rate of 1.4 m/s 2 .a. What is the true weight of the man in newtons?The true weight of the man isW = mg = (60 kg)(9.8 m/s 2 ) = 588 N.b. What is the net force acting on the man required to produce the acceleration?The net force acting on the man is∑F = ma = (60 kg)(1.4 m/s 2 ) = 84 Nin the downwards direction.c. What is the force exerted on the man’s feet by the floor of the elevator?The only two forces acting on the man are (1) a downward gravitational force of magnitude W = mgexerted by the Earth and (2) an upward normal force of magnitude N exerted by the surface of thefloor of the elevator. Therefore, using Newton’s second law, N −W = −ma (the acceleration is negativebecause the elevator is accelerating in the downwards direction). Solving for N we find that(N = W −ma = m(g −a) = (60 kg) 9.8 m/s 2 −1.4 m/s 2) = 504 N;this force is directed upwards.d. What is the apparent weight of the man in newtons? (This is the weight that wouldbe read on the scale dial if the man were standing on a bathroom scale in the acceleratingelevator.)A scale simply measures the magnitude of whatever force exerted on it. In the case of a man standingon the scale, it measures the magnitude of the force exerted by the man on the scale. By Newton’s thirdlaw, this force has the same magnitude has the force exerted by the scale on the man (the normal force).Therefore, the man’s apparent weight is simply the normal force found in part c: N = 504 N.e. How would your answers to parts b through d change if the elevator were acceleratingupward with an acceleration of 1.4 m/s 2 ?In part b, the net force acting on the man would be∑F = ma = (60 kg)(1.4 m/s 2 ) = 84 Nin the upwards direction. In part c, the force exerted on the man’s feet by the floor of the elevator wouldbe(N = W +ma = m(g +a) = (60 kg) 9.8 m/s 2 +1.4 m/s 2) = 672 Nin the upwards direction. In part d, the apparent weight of the man would be 672 N.
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Homework Set 3: Solutions Due: Tuesday, September 14, 2010 ...

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