Department of Natural Sciences Clayton College & State University. Physics 1111 Quiz 5. a. Calculate the work done by each force on the crate.

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1 Clayton College & State University October, 00 Physics 1111 Quiz 5 Name SOLUTION A crate of 50.0 kg mass containing a new lab instrument is dragged by enthusiastic physics students a distance of 30.0 m along a straight, level corridor to a physics lab. The students maintain a constant pull of magnitude 00 N applied to the crate by means of a rope inclined to the horizontal at angle of 40.0 o. The velocity of the crate along the straight path is not constant. a. Calculate the work done by each force on the crate. s = 30.0 m A = 00 N = 40.0 o W A = A cos( )s = (00.N) cos(40.0 o ) (30.0 m) = 4596 J W n = n cos( )s = n cos(90.0 o ) (30.0 m) = 0 J W w = w cos( )s = w cos(90.0 o ) (30.0 m) = 0 J b. Calculate the total work done on the crate. W NET = W A + W n + W w = 4596 J + 0 J + 0 J = 4596 J c. If initially the crate was at rest, what is the final speed of the crate as it reaches the lab? W NET = ½ m V f ½ mv i V i = 0 m/s ½ mv f = W NET V f = sqrt( W NET /m) = 13.6 m/s Clayton College & State University October 9, 00

2 Physics 1111 Quiz 6 Name SOLUTION 1. At the Longbranch saloon, a full beer stein of mass.00 kg is slid horizontally (and without rotation) across the bar with an initial speed of 3.00 m/s. If the stein travels a distance 1.53 m before stopping without loosing a drop of amber liquid, find the coefficient of kinetic friction k for the stein on the bar surface. V i = 3.00 m/s V f = 0 m/s KE i = ½ m V i = ½ (.00 kg)(3.00 m/s) = 9.00 J KE f = 0 J Work Kinetic energy theorem including non-conservative forces: W nc = KE f - KE i = 0 J 9.00 J = J W nc = W friction = f k cos(180 o )s = - f k s m) = 5.88 N f k = - W nc / s = - (-9.00 J)/(1.53 f k = k n k = f k /n n = mg (to prove it you need to draw a free-body diagram for the beer stein!) k = f k /mg = (5.88 N)/ (.00 kg x 9.80 m/s) = Physics 1111 Quiz 6.1 October 13, 005 Name SOLUTION 1. The units of work are a. J. b. N-m.

3 c. kg-m /s. d. All of the above.. You throw a ball straight up. Compare the sign of the work done by gravity while the ball goes up with the sign of the work done by gravity while it goes down. a. Work is + on the way up and + on the way down. b. Work is + on the way up and - on the way down. c. Work is - on the way up and + on the way down. d. Work is - on the way up and - on the way down. 3. If the net work done on an object is positive, then the object's kinetic energy a. Decreases. b. Remains the same. c. Increases. d. Is zero. Physics 1111 Quiz 6 October 1, 005 Name SOLUTION 1. If you push twice as hard against a stationary brick wall, the amount of work you do a. Doubles. b. Is cut in half. c. Remains constant but non-zero. d. Remains constant at zero.. You throw a ball straight up. Compare the sign of the work done by gravity while the

4 ball goes up with the sign of the work done by gravity while it goes down. a. Work is + on the way up and + on the way down. b. Work is + on the way up and - on the way down. c. Work is - on the way up and + on the way down. d. Work is - on the way up and - on the way down. 3. If the net work done on an object is negative, then the object's kinetic energy a. Decreases. b. Remains the same. c. Increases. d. Is zero. Physics 1111 Quiz 5 October 9, 007 Name _SOLUTION A crate of 50.0 kg mass containing new physics textbooks is dragged by enthusiastic physics students a distance of 30.0 m along a 5 o incline to a physics lab. The students maintain a constant pull of magnitude 50 N applied to the crate by means of a rope parallel to the incline. a. Calculate the work done by each force on the crate. Ignore friction. W A = A x cos ( ) = (50 N)(30.0m) = 7500 J W w = (mg) x cos ( ) = (50.0 kg)(9.81 m/s )(30.0m)cos ( ) = -619 J W n = n x cos ( ) = 0 b. Calculate the total work done on the crate.

5 W net = W A + W w + W n = 181 J c. If initially the crate was at rest, what is the final speed of the crate as it reaches the lab? W net = K f - K i = ½ m V f V f = W net /m V f = ( W net /m) 1/ V f = 7.16 m/s Clayton College & State University Physics 1111 Quiz 5 February 8, 005 Name SOLUTION Farmer Benton hitches his tractor to a sled loaded with firewood and pulls it a distance of 0.0 m along level ground. The total weight of the sled and load is 14,700 N. The tractor exerts a constant 5000-N force at an angle of 36.9 o to the horizontal, as shown below. There is a 3500-N friction force opposing the sled s motion. Find the work done by each force acting on the sled and the total work done by all the forces. W w = w x cos ( w ) = (14,700 N) (0.0 m) cos (90 o ) = 0 J W n = n x cos ( n ) = (14,700 N) (0.0 m) cos (90 o ) = 0 J W A = A x cos ( A ) = (5000 N) (0.0 m) cos (36.9 o ) = 79,968 J W f = f k x cos ( f ) = (3500) (0.0 m) cos (180 o ) = -70,000 J W tot = W w + W n + W A + W f = 9,968 J March 1, 006

6 Physics 1111 Quiz 5 Name SOLUTION 1. An 88-g arrow is fired from a bow whose string exerts an average force of 110 N on the arrow over a distance of 78 cm. What is the speed of the arrow as it leaves the bow? W = F x cos( = (110 N)(0.780 m)cos(0 o ) = 85.8 J W = ½ m V V = W/ m V = ( W/ m) 1/ V = ( ()85.8 J/ ( kg)) 1/ = 44. m/s. A 7.0-kg monkey swings from one branch to another 1. m higher. What is the change in potential energy? y = y f y i = 1.0 m U = m g y f m g y i = m g ( y) = (7.00 kg) (9.81 m/s )(1.0 m) = 8.4 J Physics 1111 Quiz 6 March 15, 006 Name SOLUTION 1. A skier traveling 1.0 m s reaches the foot of a steady upward 18.0º incline and glides 1. m up along this slope before coming to rest. What was the average magnitude of friction force acting on the skier? W nc = E f - E i

7 E i = K i + U gi = ½ m V i (U gi = 0) E f = K f + U gf = m g y f (K f = 0) Where y f = d sin ( ) = (1. m) sin (18.0º) = 3.77 m W nc = W friction = f k cos(180 o )s = - f k s m) = 5.88 N f k = - W nc / s = - (-9.00 J)/(1.53 Clayton College & State University February 8, 007 Physics 1111 Quiz 6 Name _SOLUTION 1. The quantity mgy is a. The kinetic energy of the object. b. The gravitational potential energy of the object. c. The work done on the object by the force. d. The power supplied to the object by the force.. An acorn falls from a tree. Compare its kinetic energy K, to its potential energy U. a. K increases and U decreases. b. K decreases and U decreases. c. K increases and U increases. d. K decreases and U increases. 3. A 30-N box is pulled 6.0 m up along a 37 inclined plane. What is the work done by the weight (gravitational force) of the box?

8 a J b J W g = U gi - U gf = mgy i mgy f = 0 (30.0 N)(6.00 m)(sin 37 ) = -108 J c J d J Physics 1111 Quiz 6 February 7, 008 Name SOLUTION You are pushing a 7.00 kg box up a rough wall with a force of 185 N directed at 30.0 o to the horizontal. The magnitude of the friction force acting on the box is 11.0 N. a. Calculate the work done by each force on the box after you pushed it through m. W = F x cos ( ) W A = A x cos( ) = (185.N) (0.750 m) cos(60.0 o ) = 69.4 J W n = n x cos( ) = n (0.750) cos(90.0 o ) = 0 J W w = w x cos( ) = (7.00 kg)(9.81 m/s ) (0.750 m) cos(180 o )= J

9 W f = f x cos( ) = (11.0 N)(0.750 m) cos(180 o )= J b. Calculate the total work done on the box. W NET = W A + W n + W w + W f = 69.4 J + 0 J J J = 9.65 J c. If initially the box was at rest, what is the final speed of the box after you pushed it through m? W NET = ½ m V f ½ mv i V i = 0 m/s ½ mv f = W NET V f = sqrt( W NET /m) = 1.66 m/s Physics 1111 Quiz 7 July 6, 005 Name SOLUTION A 145-g baseball is dropped from a tree 13.0 m above the ground. (a) With what speed would it hit the ground if air resistance could be ignored? TE i = ½ mv i + m g y i = m g y i, since V i = 0 TE f = ½ mv f + m g y f = ½ mv f, since y f = 0 TE i = TE f m g y i = ½ mv f g y i = ½ V f V f = ( g y i ) 1/

10 V f = ( (9.80 m/s ) (13.0 m) ) 1/ V f = m/s (b) If it actually hits the ground with a speed of 8.00 m s, what is the average force of air resistance exerted on it? W net = TE f - TE i = ½ mv f - m g y i W net = f x cos (180 o ) = -f x -f x = ½ mv f - m g y i f = (- ½ mv f +m g y i )/ x f = 1.06 N Physics 1111 Quiz 7 July, 007 Name SOLUTION A skier with a mass of 70.0 kg glides certain distance up the slope (30.0 o above the horizontal) before coming to rest. If the speed of the skier at the bottom of the slope was 15.0 m/s and friction can be neglected, how far did he move along the slope? E i = m g y i + ½ mv i corresponds to U g = 0) = ½ mv i (We assume that the bottom of the slope E f = m g y f + ½ mv f = m g y f (since V f = 0) E i = E i ½ mv i = m g y f ½ V i = g y f (½ V i )/g = y f

11 y f = 11.5 m d = y f / sin(30.0 o ) = 3.0 m

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