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

Homework Set 3: SolutionsDue: Tuesday, September 14, 2010Chapter 4(Q6) 5 points(Q9) 6 points(Q16) 6 pointsTwo equal forces act on two different objects, one of which has a mass ten times as large asthe other. Will the more massive object have a larger acceleration, an equal acceleration,or a smaller acceleration than the less massive object? Explain.Using Newton’s second law ∑ −→ F = m−→ a, the more massive object will experience a smaller accelerationthan the less massive one. This is assuming, of course, that there are no additional forces acting on theobjects.Is it possible that the object pictured in question 8 is moving, given the fact that the twoforces acting on it are equal in size but opposite in direction? Explain.Since the net force acting on the object is zero, we can conclude that the object is not accelerating.However, it could still be moving at a constant velocity. Therefore, it is possible that the object ismoving.The gravitational force acting on a lead ball is much larger than that acting on a woodenball of the same size. When both are dropped, does the lead ball accelerate at the samerate as the wooden ball? Explain, using Newton’s second law of motion.Ignoring the effects of air resistance, both the lead ball and the wooden ball should fall at the same rate.Because W = mg, the gravitational force acting on the lead ball is much larger than that acting on thewooden ball (the mass of the lead ball is greater than the mass of the wooden ball). However, usingNewton’s second law, for an object of mass m in free-fall:−→ a =∑ −→F/m = m−→ g/m =−→ g.Therefore, the mass of each ball does not affect its free-fall acceleration.(Q20) 12 points A boy sits at rest on the floor. What two vertical forces act upon the boy? Do thesetwo forces constitute an action/reaction pair as defined by Newton’s third law of motion?Explain.The two forces acting on the boy are: (1) a downward gravitational force exerted by the Earth on theboy and (2) an upward normal force exerted by the surface of the floor on the boy. These two forces donot constitute an “action/reaction” pair as defined by Newton’s third. The actual paired forces to thetwo forces acting on the boy are (1) an upward gravitational force exerted by the boy on the Earth and(2) a downward normal force exerted by the boy on the surface of the floor.(Q22) 10 points It is difficult to stop a car on an icy road surface. Is it also difficult to accelerate a car onthis same icy road? Explain.On an icy road surface, it is difficult to accelerate in the forward direction. This is because frictionbetween the tires and the surface of the road are responsible for a car’s acceleration (speeding up, slowingdown, and changing direction).(Q32) 10 points If you get into an elevator on the top floor of a large building and the elevator begins toaccelerate downward, will the normal force pushing up on your feet be greater than, equalto, ore less than the force of gravity pulling downward on you? Explain.If you are in an elevator, the only two forces acting on you are (1) a downward gravitational force exertedby the Earth and (2) an upward normal force exerted by the surface of the floor of the elevator. If theelevator begins to accelerate downward, the net force acting on you must also be downward (Newton’ssecond law). Therefore, the downward gravitational force must be larger than the upward normal force.

(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.