05 homework ALIBHAI, ZAHID Latest unpenalized work: Feb 19 2007 Monday 04:00 (after this date you can not make a perfect score). Work cutoff: Feb 21 2007, 4:00 am. Question 1 part 1 of 4 10 points A box of mass m with an initial velocity of v0 slides down a plane, inclined at with respect to the horizontal. The coefficient of kinetic friction is . The box stops after sliding a distance x. How much work is done by friction?
2 m v0 cos 1. W = - 2 ( sin - cos )
1
2. W = - m g x sin 3. W = - m v0 2 ( - tan )
4. W = - m g x tan 5. W = -m g ( cos + sin ) 6. W = - m g x cos 7. W = - m v0 2 ( + tan )
m
v0
k
8. W = -m g ( cos - sin ) 9. W = 0
How far does the box slide? 1. x = 2. x = 3. x = 4. x = 5. x = 6. x = 7. x = 8. x = 9. x = 10. x =
2 v0 2 g ( cos + sin ) 2 v0 g (sin - cos ) 2 v0 2 g ( sin + cos ) 2 v0 g ( sin - 2 cos ) 2 v0 2 g sin 2 v0 2 g (sin - cos ) 2 v0 2 g cos 2 v0 g ( sin + cos ) 2 v0 2 g ( sin - cos ) 2 v0 2 g ( cos - sin )
10. W = -
2 m v0 2 ( + tan )
Question 3 part 3 of 4 10 points How much work is done by the normal force? 1. W = - m v0 2 ( + tan )
2. W = - m g x cos 3. W = -m g ( cos + sin ) 4. W = 0 5. W = -
2 m v0 2 ( + tan )
6. W = - m g x sin 7. W = - m g x tan
2 m v0 cos 2 ( sin - cos ) m v0 9. W = - 2 ( - tan )
8. W = -
Question 2 part 2 of 4 10 points
10. W = -m g ( cos - sin )
homework 05 ALIBHAI, ZAHID Question 4 part 4 of 4 10 points How much work is done by gravity? m v0 tan m v0 2. W = 2 tan 1. W = 3. W = m g x cos 4. W = m g (cos + sin ) 5. W =
2 m v0 tan
2
Question 6 part 2 of 3 10 points For a force of F = 64 N, find the magnitude of the frictional force. Answer in units of N. Question 7 part 3 of 3 10 points Find the force F needed to keep the block moving up with a constant velocity. Answer in units of N. Question 8 part 1 of 2 10 points A 4 kg object is given a displacement s = ^ (3 m)^ + (3 m) ^ + (-6 m) k along a straight i line. During the displacement, a constant ^ force F = (4 N)^ + (-2 N) ^ + (3 N) k acts on i the object. Find the work done by F for this displacement. Answer in units of J. Question 9 part 2 of 2 10 points Find the component of F in the direction of this displacement. Answer in units of N. Question 10 part 1 of 1 10 points A 109 g bullet is fired from a rifle having a barrel 0.554 m long. Assuming the origin is placed where the bullet begins to move, the force exerted on the bullet by the expanding gas is F = a + b x - c x2 , where a = 21500 N, b = 8090 N/m, c = 13700 N/m2 , with x in meters. Determine the work done by the gas on the bullet as the bullet travels the length of the barrel. Answer in units of J. Question 11 part 1 of 3 10 points
6. W = m g (cos - sin ) 7. W = m g x tan 8. W =
2 m v0 2 tan
9. W = m g x sin 10. W = 0 Question 5 part 1 of 3 10 points As shown in the figure, a block of mass 2.2 kg is pushed up against the vertical wall by a force of 64 N acting at 48 to the ceiling. The coefficient of kinetic friction between the block and the wall is 0.63. The acceleration of gravity is 9.8 m/s2 . k = 0.63
2.2 kg
48
Find the work done by this force in moving the block upward by a distance 1.6 m. Answer in units of J.
F
homework 05 ALIBHAI, ZAHID A 15.7 kg block is dragged over a rough, horizontal surface by a constant force of 95.1 N acting at an angle of 32 above the horizontal. The block is displaced 39.2 m, and the coefficient of kinetic friction is 0.175. The acceleration of gravity is 9.8 m/s2 .
.1 95 N
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Answer in units of J. Question 15 part 1 of 2 10 points A 4.87 g bullet moving at 693.8 m/s penetrates a tree trunk to a depth of 4.39 cm. a) Use work and energy considerations to find the magnitude of the force that stops the bullet. Answer in units of N. Question 16 part 2 of 2 10 points b) Assuming that the frictional force is constant, determine how much time elapses between the moment the bullet the enters tree and the moment the bullets stops moving. Answer in units of s. Question 17 part 1 of 2 10 points Given: A block is pulled at a constant speed along a horizontal surface by a rope with tension FT . The rope is inclined at an angle from the horizontal as shown in the figure below. The block moves a distance 15 m along the horizontal surface. The acceleration of gravity is 9.8 m/s2 .
32
15.7 kg = 0.175 Find the work done by the 95.1 N force. Answer in units of J. Question 12 part 2 of 3 10 points Find the work done by the force of friction. Answer in units of J. Question 13 part 3 of 3 10 points If the block was originally at rest, determine its final speed. Answer in units of m/s. Question 14 part 1 of 1 10 points A 0.3 kg mass is initially at rest and is free to move with negligible friction along the xaxis. The figure below shows the value of an applied force as a function of time. 5 4 Force ( N ) 3 2 1 0 0
FT
25 kg
28
= 0.66 What is the net work done by all the forces acting on the block? 1. Wnet = 0 2. Wnet = -3675 J 3. Wnet = 1795.43 J 4. Wnet = -954.649 J 5. Wnet = -2425.5 J
-1 1 2 3 4 5 6 7 8 9 10 time ( s ) t Calculate the kinetic energy K of the mass when it reaches 5 s.
homework 05 ALIBHAI, ZAHID 6. Wnet = 2425.5 J 7. Wnet = 3675 J Question 18 part 2 of 2 10 points Suppose that the force on the block FT is increased so that the block accelerates (starting from rest) at a = 1.7 m/s2 . Now what is the net work done by all the forces acting on the block as it moves a distance of 15 m? 1. Wnet = -1795.43 J 2. Wnet = 1795.43 J 3. Wnet = 637.5 J 4. Wnet = 2267.33 J 5. Wnet = 1323.53 J 6. Wnet = -637.5 J Question 19 part 1 of 1 10 points A particle of mass m moves along the x axis. Its position varies with time according to x = (5 m/s3 ) t3 + (-9 m/s2 ) t2 . What is the work done by the force from t = 0 to t = t1 ? 1. W = 1 m t2 (81 m/s6 ) t2 1 1 2 +(-144 m/s5 ) t1 + (64 m/s4 ) 18 m t2 1 (4 m/s
6
4
+(-480 m/s5 ) t1 + (256 m/s4 ) 5. W = 9 m t2 (9 m/s6 ) t2 1 1 2 5 +(-24 m/s ) t1 + (16 m/s4 ) 9 m t2 (25 m/s6 ) t2 1 1 2 +(-20 m/s5 ) t1 + (4 m/s4 ) 9 m t2 (25 m/s6 ) t2 1 1 2 +(-60 m/s5 ) t1 + (36 m/s4 )
6. W =
7. W =
8. W = 18 m t2 (16 m/s6 ) t2 1 1 +(-24 m/s5 ) t1 + (9 m/s4 ) 9. W = 9 m t2 (49 m/s6 ) t2 1 1 2 +(-28 m/s5 ) t1 + (4 m/s4 ) 1 m t2 (729 m/s6 ) t2 1 1 2 +(-432 m/s5 ) t1 + (64 m/s4 ) Question 20 part 1 of 1 10 points A block of mass m is pushed a distance D up an inclined plane by a horizontal force F . The plane is inclined at an angle with respect to the horizontal. The block starts from rest and the coefficient of kinetic friction is k .
10. W =
D
F
m k
2. W =
) t2 1 If N is the normal force, the final speed of the block is given by 1. v = 2. v = 2 (F cos - k N ) D m 2 (F sin + k N ) D m
+(-12 m/s5 ) t1 + (9 m/s4 ) 9 3. W = m t2 (49 m/s6 ) t2 1 1 2 +(-84 m/s5 ) t1 + (36 m/s4 ) 4. W = 1 m t2 (225 m/s6 ) t2 1 1 2
homework 05 ALIBHAI, ZAHID 2 (F cos - m g sin + k N ) D m 2 (F cos + m g sin ) D m 2 (F cos - m g sin - k N ) D m 2 (F cos + m g sin - k N ) D m 2 (F cos - m g sin ) D m 2 (F sin - k N ) D m Question 21 part 1 of 3 10 points A 2.94 kg ball is attached to a ceiling by a 1.15 m long string. The height of the room is 3.69 m. The acceleration of gravity is 9.81 m/s2 . What is the gravitational potential energy associated with the ball relative to a) the ceiling? Answer in units of J. Question 22 part 2 of 3 10 points b) the floor? Answer in units of J. Question 23 part 3 of 3 10 points c) a point at the same elevation as the ball? Answer in units of J. Question 24 part 1 of 1 10 points Given: G = 6.67259 10-11 N m2 /kg2 Two spheres have equal radius R = 10.1 m and mass m = 1400 kg. Their centers are separated a distance 4R. The spheres are released from rest. What will be their speed when they collide? Answer in units of m/s.
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3. v = 4. v = 5. v = 6. v = 7. v = 8. v =